Download UNIX Tutorial One - Computer Science Departmrnt

SREE CHAITANAYA Engineering College
Department of Computer Science
Engineering
Database Management Systems
Lab Manual [CS 05157]
II B Tech – I Semester [Branch: CSE]
INSTRUCTORS
Document NO:
:
Mr. Rishi Sayal, HOD
Mr. Saidulu, Asst. Prof.
Date of Issue:
Compiled by
Date of revision
Verified by
GNEC/CSE/LAB
MANUAL/DBMS
PROGRAMMER :
jaipal
Authorized by
DBMS Lab Manual
INDEX
S. No
1
Lab Objective
2
Introduction About Lab
3
Standard Operating Procedure –
SOP
3
Guidelines to Students
4
5
Description about SQL, PL/SQL
statements and D2K.
List of Lab Exercises
4.1 Syllabus Programs (JNTU)
4.2 Additional Programs
6
Background Theory
7
Solutions for Programs
8
PL-Sql Programs
9
Additional Viva Question
10
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Contents
References
Page. no
3
4
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DBMS Lab Manual
LAB OBJECTIVE
Upon successful completion of this Lab the student will be able to:
 Creating database objects
 Modifying database objects
 Manipulating the data
 Retrieving the data from the database server
 Performing database operations in a procedural manner using pl/sql
 Performing database operations (create, update, modify, retrieve,
etc.,) using front-end tools like D2K.
 Design and Develop applications like banking, reservation system,
etc.,
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INTRODUCTION ABOUT LAB
There are 66 systems ( Compaq Presario ) installed in this
Lab. Their configurations are as follows :
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Mouse
:
Optical Mouse
Network Interface card :
Present
Software
 All systems are configured in DUAL BOOT mode i.e., Students
can boot from Windows XP or Linux as per their lab requirement.
This is very useful for students because they are familiar with
different Operating
Systems so that they can execute their
programs in different programming environments.
 Each student has a separate login for database access
Oracle 9i client version is installed in all systems. On the server,
account for each student has been created.
This is very useful because students can save their work
(scenarios’, pl / sql programs, data related projects, etc) in their
own accounts. Each student work is safe and secure from other
students.
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 Latest Technologies like DOT NET and J2EE are installed in
some systems. Before submitting their final project, they can
start doing mini project from 2nd year onwards.
 MASM (Macro Assembler) is installed in all the systems
Students can execute their assembly language programs using
MASM. MASM is very useful students because when they execute
their programs they can see contents of Processor Registers
and how each instruction is being executed in the CPU.
 Rational Rose Software is installed in some systems
Using this software, students can depict UML diagrams of their
projects.
 Softwares installed : C, C++, JDK1.5, MASM, OFFICE-XP, J2EE
and DOT NET, Rational Rose.
 Systems are provided for students in the 1:1 ratio.
 Systems are assigned numbers and same
allotted for students when they do the lab.
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DBMS Lab Manual
STANDARD OPERATING PROCEDURE – SOP
a) Explanation on today’s experiment by the concerned faculty using
OHP/PPT covering the following aspects:
25 mins.
1) Name of the experiment/Aim
2) Software/Hardware required
3) Algorithm
4) Test Data
1) Valid data sets
2) Limiting value sets
3) Invalid data sets
b) Writing of source program by the students
min.
25
c) Compiling and execution of the program
Writing of the experiment in the Observation Book:
The students will write the today’s experiment in the Observation book
as per the following format:
a) Name of the experiment/Aim
b) Software/Hardware required
c) Algorithm
d) Source Program
e) Test Data
a. Valid data sets
b. Limiting value sets
c. Invalid data sets
f) Results for different data sets
g) Viva-Voc Questions and Answers
h) Errors observed (if any) during compilation/execution
i) Signature of the Faculty
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DBMS Lab Manual
Guidelines to Students
 Equipment in the lab for the use of student community. Students
need to maintain a proper decorum in the computer lab. Students
must use the equipment with care. Any damage is caused is
punishable.
 Students are required to carry their observation / programs book
with completed exercises while entering the lab.
 Students are supposed to occupy the machines allotted to them
and are not supposed to talk or make noise in the lab. The
allocation is put up on the lab notice board.
 Lab can be used in free time / lunch hours by the students who
need to use the systems should take prior permission from the
lab in-charge.
 Lab records need to be submitted on or before date of
submission.
 Students are not supposed to use floppy disks
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How to Write and execute sql, pl/sql commands/programs:
1). Open your oracle application by the following navigation
Start->all programs->oracle orahome.->application
development->sql.
2). You will be asked for user name, pass word and host string
You have to enter user name, pass word and host string as given
by the administrator. It will be different from one user to another user.
3). Upon successful login you will get SQL prompt (SQL>).
In two ways you can write your programs:
a). directly at SQL prompt
b). or in sql editor.
If you type your programs at sql prompt then screen will look like follow:
SQL> SELECT ename,empno,
2
sal from
3
emp;
where 2 and 3 are the line numbers and rest is the command /program……
to execute above program/command you have to press ‘/’ then enter.
Here editing the program is somewhat difficult; if you want to edit the
previous command then you have to open sql editor (by default it displays
the sql buffer contents). By giving ‘ed’ at sql prompt.(this is what I
mentioned as a second method to type/enter the program).
in the sql editor you can do all the formatting/editing/file operations directly
by selecting menu options provided by it.
To execute the program which saved; do the following
SQL> @ programname.sql
Or
SQL> Run programname.sql
Then press ‘\’ key and enter.
This how we can write, edit and execute the sql command and programs.
Always you have to save your programs in your own logins.
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DBMS Lab Manual
List of Lab Exercises
Syllabus Programs (JNTU)
S. No
1
Name of the program
Database Schema for a customer-sale scenario
Customer(Cust id : integer, cust_name: string)
Item(item_id: integer, item_name: string, price: integer)
Sale(bill_no: integer, bill_data: date, cust_id: integer,
item_id: integer, qty_sold: integer)
For the above schema, perform the following—
a) Create the tables with the appropriate integrity constraints
b) Insert around 10 records in each of the tables
c) List all the bills for the current date with the customer names
and item numbers
d)
List the total Bill details with the quantity sold, price of the
item and the final amount
e)
List the details of the customer who have bought a
product which has a price>200
f)
Give a count of how many products have been bought by
each customer
g)
Give a list of products bought by a customer having
cust_id as 5
h)
List the item details which are sold as of today
i)
Create a view which lists out the bill_no, bill_date,
cust_id, item_id, price, qty_sold, amount
Create a view which lists the daily sales date wise for the last
one week
2
Database Schema for a Student Library scenario
Student(Stud_no : integer, Stud_name: string)
Membership(Mem_no: integer, Stud_no: integer)
Book(book_no: integer, book_name:string, author: string)
Iss_rec(iss_no:integer, iss_date: date, Mem_no: integer,
book_no: integer)
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For the above schema, perform the following—
a)
Create the tables with the appropriate integrity
constraints
b)
Insert around 10 records in each of the tables
c)
List all the student names with their membership numbers
d)
List all the issues for the current date with student and
Book names
e)
List the details of students who borrowed book whose
author is CJDATE
f)
Give a count of how many books have been bought by
each student
g)
Give a list of books taken by student with stud_no as 5
h)
List the book details which are issued as of today
i)
Create a view which lists out the iss_no, iss _date,
stud_name, book name
j)
Create a view which lists the daily issues-date wise for the
last one week
3
Database Schema for a Employee-pay scenario
employee(emp_id : integer, emp_name: string)
department(dept_id: integer, dept_name:string)
paydetails(emp_id : integer, dept_id: integer, basic: integer,
deductions: integer, additions: integer, DOJ: date)
payroll(emp_id : integer, pay_date: date)
For the above schema, perform the following—
a)
Create the tables with the appropriate integrity
constraints
b)
Insert around 10 records in each of the tables
c)
List the employee details department wise
d)
List all the employee names who joined after particular
date
e)
List
the details of employees whose basic salary is
between 10,000 and 20,000
f)
Give a count of how many employees are working in each
department
g)
Give a names of the employees whose netsalary>10,000
h)
List the details for an employee_id=5
i)
Create a view which lists out the emp_name, department,
basic, dedeuctions, netsalary
j)
Create a view which lists the emp_name and his netsalary
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4
Database Schema for a Video Library scenario
Customer(cust_no: integer,cust_name: string)
Membership(Mem_no: integer, cust_no: integer)
Cassette(cass_no:integer,
cass_name:string,
Language:
String)
Iss_rec(iss_no: integer, iss_date: date, mem_no: integer,
cass_no: integer)
For the above schema, perform the following—
a)
Create the tables with the appropriate integrity
constraints
b)
Insert around 10 records in each of the tables
c)
List all the customer names with their membership
numbers
d)
List all the issues for the current date with the customer
names and cassette names
e)
List the details of the customer who has borrowed the
cassette whose title is “ The Legend”
f)
Give a count of how many cassettes have been borrowed
by each customer
g)
Give a list of book which has been taken by the student
with mem_no as 5
h)
List the cassettes issues for today
i)
Create a view which lists outs the iss_no, iss_date,
cust_name, cass_name
j)
Create a view which lists issues-date wise for the last one
week
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Database Schema for a student-Lab scenario
Student(stud_no: integer, stud_name: string, class: string)
Class(class: string, descrip: string)
Lab(mach_no: integer, Lab_no: integer, description: String)
Allotment(Stud_no: Integer, mach_no: integer, dayof
week: string)
For the above schema, perform the following—
a)
Create the tables with the appropriate integrity
constraints
b)
Insert around 10 records in each of the tables
c)
List all the machine allotments with the student names,
lab and machine numbers
d)
List the total number of lab allotments day wise
e)
Give a count of how many machines have been allocated
to the ‘CSIT’ class
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f)
Give a machine allotment etails of the stud_no 5 with his
personal and class details
g)
Count for how many machines have been allocatedin
Lab_no 1 for the day of the week as “Monday”
h)
How many students class wise have allocated machines in
the labs
i)
Create a view which lists out the stud_no, stud_name,
mach_no, lab_no, dayofweek
j)
Create a view which lists the machine allotment details for
“Thursday”.
Write a program to find largest number from the given three
numbers.
Simple programs using loop, while and for iterative control
statement.
Write a program to check whether the given number is
Armstrong or not
Write a program to generate all prime numbers below 100.
Write a program to demonstrate the GOTO statement.
Write a program to demonstrate %type and %rowtype
attributes
Write a program to demonstrate predefined exceptions
Write a program to demonstrate user defined exceptions
Create a cursor, which displays all employee numbers and
names from the EMP table.
Create a cursor, which update the salaries of all employees as
per the given data.
Create a cursor, which displays names of employees having
salary > 50000.
Create a procedure to find reverse of a given number
Create a procedure to update the salaries of all employees as
per the given data
Create a procedure to demonstrate IN, OUT and INOUT
parameters
Create a function to check whether given string is palindrome or
not.
Create a function to find sum of salaries of all employees
working in depart number 10.
Create a trigger before/after update on employee table for each
row/statement.
Create a trigger before/after delete on employee table for each
row/statement.
Create a trigger before/after insert on employee table for each
row/statement.
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Create a Form to display employee details using SQL
Create a Report to generate all employee annual salaries….
Additional Programs
S. No
1
2
3
4
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Name of the Program
Create a form using Forms 6i to display Employee table
data.
Create a Master/details relationship form which
perform Add New, Search, Delete, Save and Update
on the records
Generate a report to calculate employee’s salaries
department wise from employee table.
Create a Report to generate the details of employee
table including sum and average salaries department
wise.
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DBMS Lab Manual
Background Theory
Oracle workgroup or server is the largest selling RDBMS product.it is
estimated that the combined sales of both these oracle database product
account for aroud 80% of the RDBMSsystems sold worldwide.
These products are constantly undergoing change and evolving. The natural
language of this RDBMS product is ANSI SQL,PL/SQL a superset of ANSI
SQL.oracle 8i and 9i also under stand SQLJ.
Oracle corp has also incorporated a full-fledged java virtual machine into its
database engine.since both executable share the same memory space the
JVM can communicate With the database engine with ease and has direct
access to oracle tables and their data.
SQL is structure query language.SQL contains different data types those are
1. char(size)
2. varchar2(size)
3. date
4. number(p,s)
5. long
6. raw/long raw
Different types of commands in SQL:
A).
B).
C).
D).
DDL commands: - To create a database objects
DML commands: - To manipulate data of a database objects
DQL command: - To retrieve the data from a database.
DCL/DTL commands: - To control the data of a database…
DDL commands:
1. The Create Table Command: - it defines each column of the table
uniquely. Each column has minimum of three attributes, a name , data type
and size.
Syntax:
Create table <table
<datatype><size>));
name>
(<col1>
<datatype>(<size>),<col2>
Ex:
create table emp(empno number(4) primary key, ename char(10));
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2. Modifying the structure of tables.
a)add new columns
Syntax:
Alter
table
<tablename>
col>datatype(size));
Ex:
add(<new
col><datatype(size),<new
alter table emp add(sal number(7,2));
3. Dropping a column from a table.
Syntax:
Alter table <tablename> drop column <col>;
Ex:
alter table emp drop column sal;
4. Modifying existing columns.
Syntax:
Alter table <tablename> modify(<col><newdatatype>(<newsize>));
Ex:
alter table emp modify(ename varchar2(15));
5. Renaming the tables
Syntax:
Rename <oldtable> to <new table>;
Ex:
rename emp to emp1;
6. truncating the tables.
Syntax:
Truncate table <tablename>;
Ex:
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trunc table emp1;
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7. Destroying tables.
Syntax:
Drop table <tablename>;
Ex:
drop table emp;
DML commands:
8. Inserting Data into Tables: - once a table is created the most natural
thing to do is load this table with data to be manipulated later.
Syntax:
insert into <tablename> (<col1>,<col2>) values(<exp>,<exp>);
9. Delete operations.
a) remove all rows
Syntax:
delete from <tablename>;
b) removal of a specified row/s
Syntax:
delete from <tablename> where <condition>;
10. Updating the contents of a table.
a) updating all rows
Syntax:
Update <tablename> set <col>=<exp>,<col>=<exp>;
b) updating seleted records.
Syntax:
Update
<tablename>
where <condition>;
set
<col>=<exp>,<col>=<exp>
11. Types of data constrains.
a) not null constraint at column level.
Syntax:
<col><datatype>(size)not null
b) unique constraint
Syntax:
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Unique constraint at column level.
<col><datatype>(size)unique;
c) unique constraint at table level:
Syntax:
Create table
tablename(col=format,col=format,unique(<col1>,<col2>);
d) primary key constraint at column level
Syntax:
<col><datatype>(size)primary key;
e) primary key constraint at table level.
Syntax:
Create table tablename(col=format,col=format
primary key(col1>,<col2>);
f) foreign key constraint at column level.
Syntax:
<col><datatype>(size>) references <tablename>[<col>];
g) foreign key constraint at table level
Syntax:
foreign key(<col>[,<col>])references <tablename>[(<col>,<col>)
h) check constraint
check constraint constraint at column level.
Syntax: <col><datatype>(size) check(<logical expression>)
i) check constraint constraint at table level.
Syntax: check(<logical expression>)
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DQL Commands:
12. Viewing data in the tables: - once data has been inserted into a
table, the next most logical operation would be to view what has been
inserted.
a) all rows and all columns
Syntax:
Select <col> to <col n> from tablename;
Select * from tablename;
13. Filtering table data: - while viewing data from a table, it is rare that
all the data from table will be required each time. Hence, sql must give us a
method of filtering out data that is not required data.
a) Selected columns and all rows:
Syntax:
select <col1>,<col2> from <tablename>;
b) selected rows and all columns:
Syntax:
select * from <tablename> where <condition>;
c) selected columns and selected rows
Syntax:
select <col1>,<col2> from <tablename> where<condition>;
14. Sorting data in a table.
Syntax:
Select * from <tablename> order by <col1>,<col2> <[sortorder]>;
DCL commands:
Oracle provides extensive feature in order to safeguard information stored in
its tables from unauthoraised viewing and damage.The rights that allow the
user of some or all oracle resources on the server are called privileges.
a) Grant privileges using the GRANT statement
The grant statement provides various types of access to database objects
such as tables,views and sequences and so on.
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Syntax:
GRANT <object privileges>
ON <objectname>
TO<username>
[WITH GRANT OPTION];
b) Reoke permissions using the REVOKE statement:
The REVOKE statement is used to deny the Grant given on an object.
Syntax:
REVOKE<object privilege>
ON
FROM<user name>;
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Solutions for programs (JNTU Syllabus)
1. Database Schema for a customer-sale scenario
Customer(Cust id : integer, cust_name: string)
Item(item_id: integer, item_name: string, price: integer)
Sale(bill_no: integer, bill_data: date, cust_id: integer,
integer, qty_sold: integer)
item_id:
For the above schema, perform the following—
a) Create the tables with the appropriate integrity constraints
b) Insert around 10 records in each of the tables
c) List all the bills for the current date with the customer names and item
numbers
d) List the total Bill details with the quantity sold, price of the item and
the final amount
e) List the details of the customer who have bought a product which has
a price>200
f) Give a count of how many products have been bought by each
customer
g) Give a list of products bought by a customer having cust_id as 5
h) List the item details which are sold as of today
i) Create a view which lists out the bill_no, bill_date, cust_id, item_id,
price, qty_sold, amount
j) Create a view which lists the daily sales date wise for the last one
week
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Aim: Create the tables with the appropriate integrity constraints and Insert
around 10 records in each of the tables
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
ORACLE
SQL> create table customer1 (cust_id number(5) primary key, cust_name
varchar2(15));
Output: Table created.
SQL> desc customer1;
Output:
Name
Null?
Type
----------------------------------------- -------- ---------------CUST_ID
NOT NULL
NUMBER(5)
CUST_NAME
VARCHAR2(15)
Valid Test Data
b)
SQL> insert into customer1 values(&custid,'&custname');
SQL> select * from customer1;
Output:
CUST_ID CUST_NAME
---------- --------------100 ramu
101 kamal
102 raju
103 raju sundaram
104 lawrence
SQL> create table item(item_id number(4) primary key,
item_name varchar2(15),price number(6,2));
SQL> dsec item
Output:
Name
Null?
Type
……………………………………………………………………………………………………
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Cust_id
NOT NULL
NUMBER(4)
Item_name
VARCHAR2(15)
PRICE
NUMBER(6,2)
SQL>insert into item values(&item_id,’&item_name’,&price);
SQL> select * from item;
Output:
ITEM_ID
ITEM_NAME
PRICE
……………………………………………………………………………………..
2334
geera
6.25
4532
corn soup
34.65
2124
lays chips
20
4531
setwet
99.99
2319
duracell
45.5
SQL>create table sale(bill_no number(5) primary key,bill_date date, cust_id
number(5) references customer(cust_id), item_id number(4) references
item(item_id),qty_sold number(4));
Out put: Table Created.
SQL>dsec sale
Output:
Name
Null?
Type
………………………………………………………………………………………..
BILL_NO
NOT NULL
NUMBER(4)
BILL_DATE
DATE
CUST_ID
NUMBER(5)
ITEM_ID
NUMBER(4)
QTY_SOLD
NUMBER(4)
SQL>insert into Sale values(&bill_no, ’&bill_date’,
&cust_id, &item_id, &qty_sold);
SQL>select * from sale;
Output:
BILL_NO
BILL_DATE
CUST_ID ITEM_ID
QTY_SOLD
………………………………………………………………………………………………………...
1450
04-JAN-06
100
2124
2
1451
04-JAN-06
101
2319
1
1452
04-JAN-06
103
4531
2
1453
04-JAN-06
102
2334
3
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1454
04-JAN-06
104
4532
3
c) List all the bills for the current date with the customer names and
item numbers
SQL> select c.custname, i.itemid, s.billno from customer c, item I,
sale s
where c.custid=s.custid and
s.billdate=to_char(sysdate);
CUSTNAME
------------John
ITEMID
--------5001
BILLNO
--------332
d) List the total Bill details with the quantity sold, price of the item and
the final amount
SQL> select i.price, s.qty,(i.price*s.qty) total from item I, sale s where
i.itemid=s.itemid;
PRICE
------120
20
5
10
350
QTY
----2
3
2
1
4
TOTAL
-------240
60
10
10
1400
e) List the details of the customer who have bought a product which has
a price>200
SQL> select c.custid, c.custname from customer c, sale s, item i where
i.price>200 and
c.custid=s.custid and i.itemid=s.itemid;
CUSTID
--------4
CUSTNAME
-------------duffy
f) Give a count of how many products have been bought by each
customer
SQL> select custid, count(itemid) from sale group by custid;
CUSTID
---------1
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COUNT(ITEMID)
--------------------2
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3
4
5
1
1
1
g) Give a list of products bought by a customer having cust_id as 5
SQL> select i.itemname from item i, sale s where s.custid=5 and
i.itemid-s.itemid;
ITEMNAME
-------------Pens
h) List the item details which are sold as of today
SQL> select i.itemid, i.itemname from item I, sale s where
i.itemid=s.itemid
and s.billdate=to_char(sysdate);
ITEMID
--------1234
ITEMNAME
------------pencil
i) Create a view which lists out the bill_no, bill_date, cust_id, item_id,
price, qty_sold, amount
SQL>create view cust as (select s.billno, s.billdate, c.custid, i. iitemid,
i.price, s.qty from customer c,sale s item I where
c.custid=s.custid and i.iemid=s.itemid);
view created.
SQL>select * from cust;
BILLNO BILLDATE
CUSTID ITEMID PRICE QTY
……………………………………………………………………………………………
3432
12-JAN-06
3
3244
120 2
4424
20-FEB-06
1
3456
20
3
332
13-MAR-06 1
1234
5
2
2343
10-MAR
5
5001
10
1
1331
11-MAR-06 4
76776
350 4
j) Create a view which lists the daily sales date wise for the last
one week
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Viva-Voce:
Q1. What is SQL?
Ans: Structured Query Language
2. What is database?
A database is a logically coherent collection of data with some inherent
meaning, representing some aspect of real world and which is designed,
built and populated with data for a specific purpose.
3. What is DBMS?
It is a collection of programs that enables user to create and maintain
a database. In other words it is general-purpose software that provides the
users with the processes of defining, constructing and manipulating the
database for various applications.
4. What is a Database system?
The database and DBMS software together is called as Database
system.
5.





Advantages of DBMS?
Redundancy is controlled.
Unauthorised access is restricted.
Providing multiple user interfaces.
Enforcing integrity constraints.
Providing backup and recovery.
6.






Disadvantage in File Processing System?
Data redundancy & inconsistency.
Difficult in accessing data.
Data isolation.
Data integrity.
Concurrent access is not possible.
Security Problems.
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2. Database Schema for a Student Library scenario
Student(Stud_no : integer, Stud_name: string)
Membership(Mem_no: integer, Stud_no: integer)
Book(book_no: integer, book_name:string, author: string)
Iss_rec(iss_no:integer, iss_date: date, Mem_no: integer, book_no:
integer)
For the above schema, perform the following—
a) Create the tables with the appropriate integrity constraints
b) Insert around 10 records in each of the tables
c) List all the student names with their membership numbers
d) List all the issues for the current date with student and Book names
e) List the details of students who borrowed book whose author is
CJDATE
f) Give a count of how many books have been bought by each student
g) Give a list of books taken by student with stud_no as 5
h) List the book details which are issued as of today
i) Create a view which lists out the iss_no, iss _date, stud_name, book
name
j) Create a view which lists the daily issues-date wise for the last one
week
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AIM: Create the tables with the appropriate integrity constraints
Insert around 10 records in each of the tables
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle
SQL>create table student(stud_no number(5) primary key,stud_name
varchar2(15));
SQL>desc student;
Name
Null? Type
………………………………………………………………………………………..
STUD_NO
NOT NULL NUMBER(5)
STUD_NAME
VARCAHR2(15)
Valid Test Data:
SQL>insert into student values(&stud_no,’&stud_name’);
SQL>select * from student;
STUD_NO STUD_NAME
....................................................................
508
513
518
524
534
HARISH
BALAJI
RAKESH
PAVAN
JOYCE
SQL>create table membership(mem_no number(5) primary key,stud_no
number(5) references student(stud)no));
SQL>dsec membership;
Name
Null?
Type
…………………………………………………………………………………………………….
MEM_NO
NOT NULL
NUMBER(5)
STUD_NO
NUMBER(5)
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SQL>insert into membership values(&mem_no,&stud_no);
Enter value for mem_no:5440
Enter value for stud_no:510
old 1:insert into membership values(&mem_no,&stud_no)
new 1:insert into membership values(5440,510)
insert into membership values(5440,510)
*
Errors Observed:
ERROR at line 1:
ORA-02291:integrity constraint(HARISH.SYS_C002724)violated-primary key
not found
SQL>select * from membership;
MEM_NO
STUD_NO
………………………………………………………………………..
5440
513
5441
508
5442
518
5443
534
5444
524
SQL>create table book(book_no number(5) primary key,book_name
varchar2(20),author varchar2(2));
SQL>desc book;
Name
Null? Type
………………………………………………………………………………………..
BOOK_NO
NOT NULL
NUMBER(5)
BOOK_NAME
VARCHAR2(20)
AUTHOR
VARCHAR2(20)
SQL>insert into book values(&book_no,’&book_name’,’&author’);
SQL>select * from book;
BOOK_NO
BOOK_NAME
AUTHOR
………………………………………………………………………………………………..
9123
DBMS
Rama Krishna
2342
JAVA
Robett wilkins
4523
Fearless tales
Alfred
8723
my ambition
Harish
7821
Harry Potter
JK Rowling
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SQL>create table lss_rec(iss_no number primary key,iss_date date,mem_no
number(5) references membership(mem_no),book_no number(5)
references book(book_no));
SQL>desc iss_rec;
Name
Null?
Type
………………………………………………………………………………………………………
ISS_NO
NOT NULL
NUMBER
ISS_DATE
DATE
MEM_NO
NUMBER(5)
BOOK_NO
NUMBER(5)
SQL>select * from iss_rec;
ISS_NO
ISS_DATE
MEM_NO
BOOK_NO
…………………………………………………………………………………………………
43
05-JAN-06
5443
4523
81
28-DEC-05
5441
8723
22
08-DEC-05
5440
7821
53
07-JAN-06
5442
9123
35
06-JAN-06
5444
2342
c) List all the student names with their membership numbers
SQL> select s.studname, m.memno from student s, membership m where
m.studno=s.studno;
STUDNAME MEMNO
------------- -------abhijeet
1001
arun
1002
arvind
1003
ashish
1004
ashwin
1005
d) List all the issues for the current date with student and Book names
SQL> select i.issno, s.studname, b.bookname from iss_rec I, membership
m, student s, book b
2 where i.memno=m.memno and m.studno=s.studno and
i.issdate=to_char(sysdate);
ISSNO
-------
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13
arvind
P&S
e) List the details of students who borrowed book whose author is
CJDATE
SQL> select * from student where studno in(select studno from membership
where memno in
2 (select memno from iss_rec where bookno in(select bookno from book
where author=’CJDATE’)));
STUDNO
---------505
STUDNAME
------------ashwin
f) Give a count of how many books have been bought by each student
SQL> select s.studno, count(i.bookno) from student s.membership m, book
b, 2 iss_rec I where s.studno=m.studno and b.bookno=i.bookno group by
s.studno;
STUDNO
---------501
502
503
504
505
COUNT(I.BOOKNO)
----------------------5
5
5
5
5
g) Give a list of books taken by student with stud_no as 5
SQL> select bookname from book where bookno in (select bookno from
iss_rec where
2 memno in(select memno from membership where
3 studno in(select studno from student where studno=5)));
BOOKNAME
------------NT
h) List the book details which are issued as of today
SQL> delete from book where bookno in(select bookno from iss_rec where
issdate=to_char(sysdate));
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delete from book where bookno in (select bookno from iss_rec where
issdate=to_char(sysdate))
Errors Observed:
ERROR at line 1:
ORA-02292: integrity constraint (SCOTT.SYS_C00840) violated – child
record found
i) Create a view which lists out the iss_no, iss _date, stud_name, book
name
j) Create a view which lists the daily issues-date wise for the last one
week
Viva-Vice:
1. Describe the three levels of data abstraction?
The are three levels of abstraction:
 Physical level: The lowest level of abstraction describes how data are
stored.
 Logical level: The next higher level of abstraction, describes what data
are stored in database and what relationship among those data.
 View level: The highest level of abstraction describes only part of entire
database.
2. Define the "integrity rules"
There are two Integrity rules.
 Entity Integrity: States that “Primary key cannot have NULL
value”
 Referential Integrity: States that “Foreign Key can be either a
NULL value or should be Primary Key value of other relation.
3. What is extension and intension?
Extension It is the number of tuples present in a table at any instance. This
is time dependent.
Intension -
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It is a constant value that gives the name, structure of table and
the constraints laid on it.
4. What is System R? What are its two major subsystems?
System R was designed and developed over a period of 1974-79 at
IBM San Jose Research Center. It is a prototype and its purpose was to
demonstrate that it is possible to build a Relational System that can be used
in a real life environment to solve real life problems, with performance at
least comparable to that of existing system.
Its two subsystems are
 Research Storage
 System Relational Data System.
5. How is the data structure of System R different from the relational
structure?
Unlike Relational systems in System R
 Domains are not supported
 Enforcement of candidate key uniqueness is optional
 Enforcement of entity integrity is optional
 Referential integrity is not enforced
6. What is Data Independence?
Data independence means that “the application is independent of the
storage structure and access strategy of data”. In other words, The ability to
modify the schema definition in one level should not affect the schema
definition in the next higher level.
Two types of Data Independence:
 Physical Data Independence: Modification in physical level should
not affect the logical level.
 Logical Data Independence: Modification in logical level should
affect the view level.
NOTE: Logical Data Independence is more difficult to achieve
7. What is a view? How it is related to data independence?
A view may be thought of as a virtual table, that is, a table that does
not really exist in its own right but is instead derived from one or more
underlying base table. In other words, there is no stored file that direct
represents the view instead a definition of view is stored in data dictionary.
Growth and restructuring of base tables is not reflected in views. Thus
the view can insulate users from the effects of restructuring and growth in
the database. Hence accounts for logical data independence.
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3. Database Schema for a Employee-pay scenario
employee(emp_id : integer, emp_name: string)
department(dept_id: integer, dept_name:string)
paydetails(emp_id : integer, dept_id: integer, basic: integer, deductions:
integer, additions: integer, DOJ: date)
payroll(emp_id : integer, pay_date: date)
For the above schema, perform the following—
a) Create the tables with the appropriate integrity constraints
b) Insert around 10 records in each of the tables
c) List the employee details department wise
d) List all the employee names who joined after particular date
e) List the details of employees whose basic salary is between 10,000
and 20,000
f) Give a count of how many employees are working in each department
g) Give a names of the employees whose netsalary>10,000
h) List the details for an employee_id=5
i) Create a view which lists out the emp_name, department, basic,
dedeuctions, netsalary
j) Create a view which lists the emp_name and his netsalary
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AIM: Create the tables with the appropriate integrity constraints
Insert around 10 records in each of the tables
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle
Create table employee(emp_id number(5) primary key,emp_name
varchar2(25));
SQL>desc employee;
Name
Null? Type
………………………………………………………………………………………..
EMP_ID
NOT NULL
NUMBER(5)
EMP_NAME
VARCHAR2(25)
Valid Test Data:
SQL>insert into employee values(&emp_id,’&emp_name’);
SQL>select * from employee;
EMP_ID
EMP_NAME
………………………………………………………….
10
Robert
21
Coulthard
30
Fernando Alonso
39
Kartikeyan
87
Kimmi
SQL>create table department(dept_id number(5) primary key,dept_name
varchar2(20));
SQL>desc department;
Name
Null?
Type
………………………………………………………………………………………..
DEPT_ID
NOT NULL
NUMBER(5)
DEPT_NAME
VARCHAR2(20)
SQL>insert into department values(&dept_id,’&dept_name’);
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SQL>select * from department;
DEPT_ID
DEPT_NAME
……………………………………………………………………………..
100
sales
101
accounts
102
administration
103
production
104
supervisor
SQL>create table paydetails(emp_id number(5) references
employee(emp_id),dept_id number(5) reerences department(dept_id),basic
number(7,2),deductions number(5,2),additions number(5,2),doj date);
SQL>desc paydetails;
Name
Null?
Type
………………………………………………………………………………………..
EMP_ID
NUMBER(5)
DEPT_ID
NUMBER(5)
BASIC
NUMBER(7,2)
DEDUCTIONS
NUMBER(5,2)
ADDITIONS
NUMBER(5,2)
DOJ
DATE
Different Data Sets:
SQL>insert into paydeatils values(&emp_id,&dept_id,
&basic,&deductions,&additions,&doj);
SQL>select * from paydeatils;
EMP_ID DEPT_ID BASIC DEDUCTIONS
ADDITIONS DOJ
…………………………………………………………………………………………………………………..
10
101
25023.12 43.09
71.23
08-JAN-93
21
100
10500.29 23.98
40.9
01-JAN-06
30
102
6500.5
30.54
15
06-JUL-97
39
103
9700.45
32.78
65.09
08-AUG-03
87
104
15000
97.66
154.8
24-SEP-04
SQL>create table payroll(emp_id number(5)references
employee(emp_id),pay_date date);
SQL>desc payroll;
Name
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Type
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………………………………………………………………………………………..
EMP_ID
NUMBER(5)
PAY_DATE
DATE
SQL>insert into payroll values(&emp_id,’&date’);
SQL>select * from payroll;
EMP_ID
PAY_DATE
………………………………………………………….
10
31-JAN-06
21
03-FEB-06
30
15-JAN-06
39
27-JAN-06
87
04-FEB-06
c) List the employee details department wise
SQL>select empid,deptid from paydet;
EMPID
DEPTID
…………………………
401
500
402
200
403
600
404
400
405
1200
d) List all the employee names who joined after particular date
SQL>select e,empname from employee e,paydet p where e.empid=p.empid
and p.doj>=’05-mar-06’;
EMPNAME
…………………
AVINASH
NITIN
PHALGUN
e) List the details of employees whose basic salary is between 10,000 and
20,000
sqL> Select empid,empname from employee where salary between 10000
and 20000;
EMPID
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…………………………….
402
AKHILA
403
aaaaaaaa
EMPID EMPNAME
…………………………….
AKHILA
f) Give a count of how many employees are working in each department
SQL>select count(empid),deptid from paydet group by deptid;
COUNT (EMPID)
DEPTID
………………………………………………………
1
200
1
400
1
500
1
600
1
1200
g) Give a names of the employees whose netsalary>10,000
SQL> select empname from employee where empid in(select empid from
paydet where basic-deduction>10000);
EMPNAME
………………
AVINASH
AKHILA
HARISH
NITIN
PHALGUN
h) List the details for an employee_id=5
SQL> select * from employee where empid=5;
EMPID
EMPNAME
-----------------------------------------5
Coulthard
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i)
j)
Create a view which lists out the emp_name, department, basic,
dedeuctions, netsalary
Create a view which lists the emp_name and his netsalary
Viva-Vice:
13. What is Data Model?
A collection of conceptual tools for describing data, data relationships
data semantics and constraints.
14.
What is E-R model?
This data model is based on real world that consists of basic objects
called entities and of relationship among these objects. Entities are
described in a database by a set of attributes.
15.
What is Object Oriented model?
This model is based on collection of objects. An object contains values
stored in instance variables with in the
object. An object also contains
bodies of code that operate on the object. These bodies of code are called
methods. Objects that contain same types of values and the same methods
are grouped together into classes.
16.
What is an Entity?
It is a 'thing' in the real world with an independent existence.
17.
What is an Entity type?
It is a collection (set) of entities that have same attributes.
18.
What is an Entity set?
It is a collection of all entities of particular entity type in the database.
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4. Database Schema for a Video Library scenario
Customer(cust_no: integer,cust_name: string)
Membership(Mem_no: integer, cust_no: integer)
Cassette(cass_no:integer, cass_name:string, Language: String)
Iss_rec(iss_no: integer, iss_date: date, mem_no: integer, cass_no:
integer)
For the above schema, perform the following—
a) Create the tables with the appropriate integrity constraints
b) Insert around 10 records in each of the tables
c) List all the customer names with their membership numbers
d) List all the issues for the current date with the customer names and
cassette names
e) List the details of the customer who has borrowed the cassette whose
title is “ The Legend”
f) Give a count of how many cassettes have been borrowed by each
customer
g) Give a list of book which has been taken by the student with mem_no
as 5
h) List the cassettes issues for today
i) Create a view which lists outs the iss_no, iss_date, cust_name,
cass_name
j) Create a view which lists issues-date wise for the last one week
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AIM: Create the tables with the appropriate integrity constraints
Insert around 10 records in each of the tables
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle
SQL>create table customer(cust_no number(5) primary key,cust_name
varchar2(20));
SQL>desc customer;
Name
Null?
Type
……………………………………………………………………………………………………………..
CUST_NO
NOT NULL
NUMBER(5)
CUST_NAME
VARCHAR2(20)
Valid Test Data:
SQL>insert into customer values(&cust_no,’&cust_name’);
SQL>select * from customer;
CUST_NO
CUST_NAME
……………………………………………………………….
50
scott
51
pandey
52
varshney
53
naidu
54
bhimbra
SQL>create table membership(mem_no number(5) primary key,cust_no
number(5) references customer(cust_no));
SQL>dsec membership;
Name
Null?
Type
………………………………………………………………………………………………………...
MEM_NO
NOT NULL
NUMBER(5)
CUST_NO
NUMBER(5)
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SQL>insert into memship values(&mem_no,&cust_no);
SQL>select * from memship;
MEM_NO
CUST_NO
…………………………………………………
920
50
981
51
897
52
820
53
928
54
SQL>create table cassette(cass_no number(5) primary key,
Cass_name varchar2(15),language varchar2(15));
SQL>desc cassette;
Name
Null?
Type
………………………………………………………………………………………..
CASS_NO
NOT NULL
NUMBER(5)
CASS_NAME
VARCHAR2(15)
LANGUAGE
VARCHAR2(15)
SQL>insert into cassette values(&cass_no,’&cass_name’,’&language’);
SQL>select * from cassette;
CASS_NO
CASS_NAME
LANGUAGE
………………………………………………………………………………………
1
tagore
telugu
2
the lion king
English
3
anniyan
tamil
4
indra
telugu
5
lord of rings
English
SQL>create table issu_rec(iss_no number(5) primary key,iss_date
date,mem_no number(5)references memship(mem_no),cass_no number(5)
references cassette(cass_no));
SQL>desc issu_rec;
Name
Null?
Type
………………………………………………………………………………………………………...
ISS_NO
NOT NULL
NUMBER(5)
ISS_DATE
DATE
MEM_NO
NUMBER(5)
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CASS_NO
NUMBER(5)
SQL>select * from issu_rec;
ISS_NO
ISS_DATE
MEM_NO
CASS_NO
……………………………………………………………………………………
22
07-JAN-06
920
1
23
10-JAN-00
981
2
26
10-JAN-06
897
5
3
01-JAN-06
820
4
34
31-DEC-05
928
3
c) List all the customer names with their membership numbers
SQL>select c.custname,m.memno from customer1 c,membership1 m
where c.custno=m.custno;
CUSTNAME MEMNO
……………….. ………………..
NIKHIL
51
VIVEK
52
SHRAVAN
58
VAMSI
57
SHIVA
56
d) List all the issues for the current date with the customer names and
cassette names
SQL>select i.issno,c.custname,cc.cassettename from customer1
c,membership1 m,cassette cc,issrec1 I where
i.issdate=to_char(sysdate) and c.custno=m.custno and
i.cassno=cc.cassno and i.memno=m.memno;
OutPut:
no rows selected.
e) List the details of the customer who has borrowed the cassette whose
title is “ The Legend”
f) Give a count of how many cassettes have been borrowed by each
customer
g) Give a list of book which has been taken by the student with mem_no
as 5
h) List the cassettes issues for today
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i) Create a view which lists outs the iss_no, iss_date, cust_name,
cass_name
j) Create a view which lists issues-date wise for the last one week
Viva-Vice:
19.
What is an Extension of entity type?
The collections of entities of a particular entity type are grouped
together into an entity set.
20.
What is Weak Entity set?
An entity set may not have sufficient attributes to form a primary key,
and its primary key compromises of its partial key and primary key of its
parent entity, then it is said to be Weak Entity set.
21.
What is an attribute?
It is a particular property, which describes the entity.
22.
What is a Relation Schema and a Relation?
A relation Schema denoted by R(A1, A2, …, An) is made up of the
relation name R and the list of attributes Ai that it contains. A relation is
defined as a set of tuples. Let r be the relation which contains set tuples (t1,
t2, t3, ..., tn). Each tuple is an ordered list of n-values t=(v1,v2, ..., vn).
23.
What is degree of a Relation?
It is the number of attribute of its relation schema.
24.
What is Relationship?
It is an association among two or more entities.
25.
What is Relationship set?
The collection (or set) of similar relationships.
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5.Database Schema for a student-Lab scenario
Student(stud_no: integer, stud_name: string, class: string)
Class(class: string, descrip: string)
Lab(mach_no: integer, Lab_no: integer, description: String)
Allotment(Stud_no: Integer, mach_no: integer, dayof week: string)
For the above schema, perform the following—
a) Create the tables with the appropriate integrity constraints
b) Insert around 10 records in each of the tables
c) List all the machine allotments with the student names, lab and
machine numbers
d) List the total number of lab allotments day wise
e) Give a count of how many machines have been allocated to the ‘CSIT’
class
f) Give a machine allotment etails of the stud_no 5 with his personal and
class details
g) Count for how many machines have been allocatedin Lab_no 1 for
the day of the week as “Monday”
h) How many students class wise have allocated machines in the labs
i) Create a view which lists out the stud_no, stud_name, mach_no,
lab_no, dayofweek
j) Create a view which lists the machine allotment details for “Thursday”
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AIM: Create the tables with the appropriate integrity constraints
Insert around 10 records in each of the tables
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle
SQL>create table stu(stud_no number(5) primary key,stud_nam
varchar2(20),class varchar2(20));
SQL> desc stu;
Name
STUD_NO
STUD_NAM
CLASS
null?
NOT NULL
Type
NUMBER(5)
VARCHAR2(20)
VARCHAR2(20)
Valid Data Sets:
SQL> insert into stu values(&stud_no,’&stud_nam’,’&class’);
SQL> select * from stu;
STUD_NO
39
34
18
8
24
STUD_NAM
LEON
VIKAS
MATHEW
HANSEN
ALEXIS
CLASS
CSE
CSIT
ECE
MECH
EEE
SQL> Create table class (class varchar2(20), descript varchar2(10));
SQL> Describe class;
Name
CLASS
DESCRIPT
null
type
VARCHAR2(10)
VARCHAR2(20)
SQL> create table lab(match_no number(5), lab_no number(5), description
varchar2(20));
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SQL> desc lab;
Name
null
MACH_NO
LAB_NO
DESCRIPTION
NOT NULL
type
NUMBER(5)
NUMBER(5)
VARCHAR2(20)
SQL> insert into lab values(&mach_no,&lab_no,’&description’);
SQL> select * from lab;
MATCH_NO LAB_NO
DESCRIPTION
-----------------------------------------23
7
physics
78
2
chemistry
87
1
edc
12
10
cds
8
3
java lab
SQL> create table allotment(stud_no number(5) references stu(stud_no),
match_no number(5) references lab(mach_no),
Doweek varchar2(20));
SQL> desc allotment;
Name
-------------STUD_NO
MACH_NO
DOWEEK
Null?
-------
Type
--------NUMBER(5)
NUMBER(5)
VARCHAR2(20)
SQL>select * from allotment;
STUD_NO
------------- -39
34
18
8
24
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MACH_NO
-----------23
87
78
12
12
DOWEEK
-----------sat
mon
tue
wed
thu
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c) List all the machine allotments with the student names, lab and
machine numbers
SQL>select s.studname,l.machno from student1 s,lab l,allotment a where
a.machno=l.machno and a.studno=s.studno;
STUDNAME
MACHNO
………………………………………..
ABHIJEET
1
KALYAN
22
ASHWIN
3
ARKA
4
ARVIND
5
d) List the total number of lab allotments day wise
SQL>select l.machno,l.descrip,a.day from lab l,allotment a where
a.machno=l.machno;
MACHNO
DESCRIP
DAY
……………………………………………………………………
1
UNIX
MONDAY
22
UNIX
TUESDAY
3
XP
WEDNESDAY
4
WINDOWS
THRUSDAY
5
ME
FRIDAY
e) Give a count of how many machines have been allocated to the ‘CSIT’
class
SQL>select count(machno)from allotment where studno in(select studno
from student1 where class=’CSIT’);
COUNT (MACHNO)
……………………..
1
f) Give a machine allotment etails of the stud_no 5 with his personal and
class details
SQL>select a.studno,a.machno,s.studname,s.class from allotment
a,student1 s where a.studno=s.studno and a.studno=503;
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STUDNO
MACHNO
STUDNAME
CLASS
………………………………………………………………………………………………………
503
5
ARVIND
CSE
g) Count for how many machines have been allocatedin Lab_no 1 for
the day of the week as “Monday”
h) How many students class wise have allocated machines in the labs
SQL>select count(studno) “allocated students in the labs”,class from
student1 where studno in(select studno from allotment) group by
class;
allocated students in the lab
CLASS
……………………………………………………………………………
2
CSE
1
ECE
1
EEE
1
IT
i) Create a view which lists out the stud_no, stud_name, mach_no,
lab_no, dayofweek
j) Create a view which lists the machine allotment details for “Thursday”
Viva-Vice:
26.
What is Relationship type?
Relationship type defines a set of associations or a relationship set
among a given set of entity types.
27.
What is degree of Relationship type?
It is the number of entity type participating.
25.
What is DDL (Data Definition Language)?
A data base schema is specifies by a set of definitions expressed by a
special language called DDL.
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26.
What is VDL (View Definition Language)?
It specifies user views and their mappings to the conceptual schema.
27.
What is SDL (Storage Definition Language)?
This language is to specify the internal schema. This language may
specify the mapping between two schemas.
28.
What is Data Storage - Definition Language?
The storage structures and access methods used by database system
are specified by a set of definition in a special type of DDL called data
storage-definition language.
29.
What is DML (Data Manipulation Language)?
This language that enable user to access or manipulate data as
organised by appropriate data model.
 Procedural DML or Low level: DML requires a user to specify what data
are needed and how to get those data.
 Non-Procedural DML or High level: DML requires a user to specify what
data are needed without specifying how to get those data.
31.
What is DML Compiler?
It translates DML statements in a query language into low-level
instruction that the query evaluation engine can understand.
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6). Write a program to find largest number from the
given three numbers.
Aim: To find largest number from the given three numbers.
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Algorithm:
Step 1: Declare the variable A, B, and C.
Step 2: Store the valid data.
Step 3: Compare variable A with B and A with C
Step 4: If the value stored in variable A is big, it displays “A is Big”. (IF
conditional statement should be used)
Step 5: Compare variable B with C
Step 6: If the value stored in variable B is big, it displays “B is Big”.
Step 7: other wise it displays “C is Big”
Declare
A number;
B number;
C number;
Begin
A:=&a;
B:=&b;
C:=&c;
If a > b && a> c then
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Dbms_output.put_line(‘ A is big ‘);
Else
If( b>c && b> a ) then
Dbms_output.put_line(‘ B is big ‘);
Else
Dbms_output.put_line(‘ C is big ‘);
End if;
End if;
End;
Valid Data Sets:
Enter the value of a:
1
Enter the value of b:
2
Enter the value of c:
3
OUTPUT:
C is big
Invalid Data sets :
Enter the value of a:
y
Enter the value of b:
x
Enter the value of c:
a
Output:
Invalid data types.
Viva-Vice:
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31. What is Pl-SQL ?
Procedural Language Structured Query Language
32.
What is Query evaluation engine?
It executes low-level instruction generated by compiler.
33.
What is DDL Interpreter?
It interprets DDL statements and record them in tables containing
metadata.
34.
What is Record-at-a-time?
The Low level or Procedural DML can specify and retrieve each record
from a set of records. This retrieve of a record is said to be Record-at-atime.
35.
What is Set-at-a-time or Set-oriented?
The High level or Non-procedural DML can specify and retrieve many
records in a single DML statement. This retrieve of a record is said to be Setat-a-time or Set-oriented.
36.
What is Relational Algebra?
It is procedural query language. It consists of a set of operations that
take one or two relations as input and produce a new relation.
37.
What is Relational Calculus?
It is an applied predicate calculus specifically tailored for relational
databases proposed by E.F. Codd. E.g. of languages based on it are DSL
ALPHA, QUEL.
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7). Simple programs using loop, while and for iterative
control statement.
a) To generate first 10 natural numbers using loop, while and for.
AIM: To generate first 10 natural numbers using loop, while and for.
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Algorithm:
Step
Step
Step
Step
Step
Step
Step
1:
2:
3:
4:
5:
6:
7:
Declare the variable I.
Store the valid data 1 in I.
Use LOOP statement
Display the first value.
Increment the value of I by 1 value.
check the value up to 10 no. and repeat the loop
If condition exceeds the given value 10, the loop will be
terminated.
/* using loop statement */
Declare
I number;
Begin
I:=1;
Loop
Dbms_output.put_line(I);
I:=I+1;
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Exit when I>10;
End loop;
End;
Algorithm: for WHILE loop
Step
Step
Step
Step
Step
Step
Step
1:
2:
3:
4:
5:
6:
7:
Declare the variable I.
Store the valid data 1 in I.
Use WHILE statement
Check the value of I with value 10.
if the value of I reached to 10 the loop will be terminated
otherwise display value of I
increment the next value of I using I=I+1.
/* using while */
Declare
I number;
Begin
I:=1;
While (I<=10)
loop
Dbms_output.put_line(I);
I:=I+1;
End loop;
End;
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Algorithm:
Step 1: Declare the variable I.
Step 2: Store the value 1 in var. I.
Step 3: Use For… LOOP statement
Step 4: Display the first value of I.
Step 5: Increment the value of I by 1 value.
Step 6: check the value up to 10 no. and repeat the loop
Step 7: if the loop exceeds the value 10 then the loop will be
terminated.
/* using for loop*/
Begin
For I in 1..10
loop
Dbms_output.put_line(I);
End loop;
End;
Valid Test Data:
OUTPUT
1
2
3
4
5
6
7
8
9
10
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Viva-Vice:
38. How does Tuple-oriented relational calculus differ from domainoriented relational calculus
The tuple-oriented calculus uses a tuple variables i.e., variable whose
only permitted values are tuples of that relation. E.g. QUEL
The domain-oriented calculus has domain variables i.e., variables that range
over the underlying domains instead of over relation. E.g. ILL, DEDUCE.
39.
What is normalization?
It is a process of analysing the given relation schemas based on their
Functional Dependencies (FDs) and primary key to achieve the properties
 Minimizing redundancy
 Minimizing insertion, deletion and update anomalies.
40.
What is Functional Dependency?
A Functional dependency is denoted by X
Y between two sets of
attributes X and Y that are subsets of R specifies a constraint on the possible
tuple that can form a relation state r of R. The constraint is for any two
tuples t1 and t2 in r if t1[X] = t2[X] then they have t1[Y] = t2[Y]. This
means the value of X component of a tuple uniquely determines the value of
component Y.
41. When is a functional dependency F said to be minimal?
 Every dependency in F has a single attribute for its right hand side.
 We cannot replace any dependency X
A in F with a dependency Y A
where Y is a proper subset of X and still have a set of dependency that is
equivalent to F.
 We cannot remove any dependency from F and still have set of
dependency that is equivalent to F.
42.
What is Multivalued dependency?
Multivalued dependency denoted by
X
Y specified on relation
schema R, where X and Y are both subsets of R, specifies the following
constraint on any relation r of R: if two tuples t1 and t2 exist in r such that
t1[X] = t2[X] then t3 and t4 should also exist in r with the following
properties
 t3[x] = t4[X] = t1[X] = t2[X]
 t3[Y] = t1[Y] and t4[Y] = t2[Y]
 t3[Z] = t2[Z] and t4[Z] = t1[Z]
where [Z = (R-(X U Y)) ]
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43.
What is Lossless join property?
It guarantees that the spurious tuple generation does not occur with
respect to relation schemas after decomposition.
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8. Program to check
Armstrong or not.
whether
given
number
is
AIM: to check whether given number is Armstrong or not.
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, Pl-SQL
Algorithm:
Step 1: Declare the variable N, S, D and DUP.
Step 2: Store the value in var. N and var. DUP..
Step 3: check for the value of N, which is not equal to 0.
Step 4: divide value stored in N by 10 and store it var. D. (D=n%10).
Step 5: the reminder will be multiply 3 times and store it in Var. S.
Step 6: The coefficient will be calculated using FLOOR function. And
store it in var. N.
Step 7: repeat the Steps 3, 4, 5, and 6 till loop will be terminated.
Step 8: Check whether the stored value and calculated values are same
Step 9: if both the values are same, then display “The given number is
Armstrong”
Step 10: Otherwise display “it is not Armstrong” and terminate the
loop.
Declare
N number;
S number;
D number;
Begin
N:=&n;
S:=0;
While(n!=0)
Loop
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D=n%10;
S:=s+(D*D*D);
N:=floor(n/10);
End loop;
If (DUP=S) then
DBMS_output.put_line(‘number is armstrong’);
Else
DBMS_output.put_line(‘number is not armstrong’);
End if;
End;
Test Valid Data Set:
Enter value of n
153
OUTPUT:
number is Armstrong.
Viva-Vice:
44. What is 1 NF (Normal Form)?
The domain of attribute must include only atomic (simple, indivisible) values.
45. What is Fully Functional dependency?
It is based on concept of full functional dependency. A functional dependency X Y is
full functional dependency if removal of any attribute A from X means that the dependency does
not hold any more.
46. What is 2NF?
A relation schema R is in 2NF if it is in 1NF and every non-prime attribute A in R is fully
functionally dependent on primary key.
47. What is 3NF?
A relation schema R is in 3NF if it is in 2NF and for every FD X
following is true
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 X is a Super-key of R.
 A is a prime attribute of R.
In other words, if every non prime attribute is non-transitively dependent on primary key.
48. What is BCNF (Boyce-Codd Normal Form)?
A relation schema R is in BCNF if it is in 3NF and satisfies an additional constraint that
for every FD X A, X must be a candidate key.
49. What is 4NF?
A relation schema R is said to be in 4NF if for every Multivalued dependency
Y that holds over R, one of following is true
 X is subset or equal to (or) XY = R.
 X is a super key.
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9. Write a program to generate all prime numbers
below 100.
AIM: to generate all prime numbers below 100.
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Declare
I number;
J number;
C number;
Begin
While(i<=100)
Loop
C:=0;
J:=1;
While(j<=i)
Loop
If(floor(i%j)=0) then
C:= C+1;
End if;
J:=j+1;
End loop;
If(c=2) then
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Dbms_output.put_line(i);
End if;
Endloop;
End;
Valid Test Data
OUTPUT:
2
3
5
7
11
.
.
99
Viva-Vice:
50. What is 5NF?
A Relation schema R is said to be 5NF if for every join dependency {R1, R2, ..., Rn}
that holds R, one the following is true
 Ri = R for some i.
 The join dependency is implied by the set of FD, over R in which the left side is key of R.
51. What is Domain-Key Normal Form?
A relation is said to be in DKNF if all constraints and dependencies that should hold on
the the constraint can be enforced by simply enforcing the domain constraint and key
constraint on the relation.
52. What are partial, alternate,, artificial, compound and natural key?
Partial Key:
It is a set of attributes that can uniquely identify weak entities and that are related to
same owner entity. It is sometime called as Discriminator.
Alternate Key:
All Candidate Keys excluding the Primary Key are known as Alternate Keys.
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Artificial Key:
If no obvious key, either stand alone or compound is available, then the last
resort is to simply create a key, by assigning a unique number to each record or occurrence. Then
this is known as developing an artificial key.
Compound Key:
If no single data element uniquely identifies occurrences within a construct, then
combining multiple elements to create a unique identifier for the construct is known as creating a
compound key.
Natural Key:
When one of the data elements stored within a construct is utilized as the primary
key, then it is called the natural key.
53. What is indexing and what are the different kinds of indexing?
Indexing is a technique for determining how quickly specific data can be found.
Types:
 Binary search style indexing
 B-Tree indexing
 Inverted list indexing
 Memory resident table
 Table indexing
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10. Write a
statement.
program
to
demonstrate
the
GOTO
AIM: to demonstrate the GOTO statement
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Declare
I number;
Begin
I:=1;
If(i>=0) then
GOTO here;
Else
Dbms_output.put_line( ‘ I is negative’);
End if;
<<here>>
Dbms_output.put_line( ‘ I is positive’);
End;
Valid Test Data
OUTPUT:
I is positive
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Viva-Vice:
54. What is system catalog or catalog relation? How is better known as?
A RDBMS maintains a description of all the data that it contains, information about
every relation and index that it contains. This information is stored in a collection of relations
maintained by the system called metadata. It is also called data dictionary.
55. What is meant by query optimization?
The phase that identifies an efficient execution plan for evaluating a query that has the
least estimated cost is referred to as query optimization.
56. What is join dependency and inclusion dependency?
Join Dependency:
A Join dependency is generalization of Multivalued dependency.A JD
{R1, R2, ..., Rn} is said to hold over a relation R if R1, R2, R3, ..., Rn is a lossless-join
decomposition of R . There is no set of sound and complete inference rules for JD.
Inclusion Dependency:
An Inclusion Dependency is a statement of the form that some columns of a
relation are contained in other columns. A foreign key constraint is an example of inclusion
dependency.
57. What is durability in DBMS?
Once the DBMS informs the user that a transaction has successfully completed, its effects
should persist even if the system crashes before all its changes are reflected on disk. This
property is called durability.
58. What do you mean by atomicity and aggregation?
Atomicity:
Either all actions are carried out or none are. Users should not have to worry
about the effect of incomplete transactions. DBMS ensures this by undoing the actions of
incomplete transactions.
Aggregation:
A concept which is used to model a relationship between a collection of entities
and relationships. It is used when we need to express a relationship among relationships.
59. What is a Phantom Deadlock?
In distributed deadlock detection, the delay in propagating local information might cause
the deadlock detection algorithms to identify deadlocks that do not really exist. Such situations
are called phantom deadlocks and they lead to unnecessary aborts.
60. What is a checkpoint and When does it occur?
A Checkpoint is like a snapshot of the DBMS state. By taking checkpoints, the DBMS
can reduce the amount of work to be done during restart in the event of subsequent crashes.
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11. Write a program to demonstrate %type and
%rowtype attributes
AIM: to demonstrate %type and %rowtype attributes
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Declare
My_Empno emp.empno%type;
My_Ename emp.ename%type;
My_Emprow
emp%rowtype;
No number;
Begin
No:=&no;
Select empno,ename into my_empno,my_ename from emp where
empno=no;
If(SQl%rowcount=1) then
Dbms_output.put_line(‘empno is’ || my_empno || ‘ename is ‘ ||
my_ename);
Else
Dbms_output.put_line( ‘error’);
End if;
Select * into my_emprow from emp where empno=no;
If(SQl%rowcount=1) then
Dbms_output.put_line(‘empno is’ || my_emprow.empno || ‘ename is ‘
|| my_emprow.ename);
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Else
Dbms_output.put_line( ‘error’);
End if;
End;
Valid Test Data
Enter the value for no:
7788
OUTPUT
empno is 7788 ename is vinay s.
empno is 7788 ename is vinay s.
Viva-Vice
61. What are the different phases of transaction?
Different phases are
 Analysis phase
 Redo Phase
 Undo phase
62. What do you mean by flat file database?
It is a database in which there are no programs or user access languages. It has no crossfile capabilities but is user-friendly and provides user-interface management.
63. What is "transparent DBMS"?
It is one, which keeps its Physical Structure hidden from user.
64. Brief theory of Network, Hierarchical schemas and their properties
Network schema uses a graph data structure to organize records example for such a
database management system is CTCG while a hierarchical schema uses a tree data structure
example for such a system is IMS.
65. What is a query?
A query with respect to DBMS relates to user commands that are used to interact with a
data base. The query language can be classified into data definition language and data
manipulation language.
66. What do you mean by Correlated subquery?
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Subqueries, or nested queries, are used to bring back a set of rows to be used by the
parent query. Depending on how the subquery is written, it can be executed once for the parent
query or it can be executed once for each row returned by the parent query. If the subquery is
executed for each row of the parent, this is called a correlated subquery.
A correlated subquery can be easily identified if it contains any references to the parent
subquery columns in its WHERE clause. Columns from the subquery cannot be referenced
anywhere else in the parent query. The following example demonstrates a non-correlated
subquery.
E.g. Select * From CUST Where '10/03/1990' IN (Select ODATE From ORDER Where
CUST.CNUM = ORDER.CNUM)
67. What are the primitive operations common to all record management systems?
Addition, deletion and modification.
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12. Write a
exceptions
program
to
demonstrate
predefined
AIM: to demonstrate predefined exceptions
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Declare
A number
B number;
C number;
Begin
A:=&a;
B:=&b;
C:=a/b;
Dbms_output.put_line(‘division is ‘ || C);
Exception
If (ZERO_DIVIDE) then
Dbms_output.put_line(‘b could not be zero’);
End if;
End;
Valid Test Data:
Enter the value for a:
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10
Enter the value for b:
0
OUTPUT:
b could not be zero
Viva-Vice:
68. Name the buffer in which all the commands that are typed in are stored
‘Edit’ Buffer
69. What are the unary operations in Relational Algebra?
PROJECTION and SELECTION.
70. Are the resulting relations of PRODUCT and JOIN operation the same?
No.
PRODUCT: Concatenation of every row in one relation with every row in another.
JOIN: Concatenation of rows from one relation and related rows from another.
71. What is RDBMS KERNEL?
Two important pieces of RDBMS architecture are the kernel, which is the software, and
the data dictionary, which consists of the system-level data structures used by the kernel to
manage the database
You might think of an RDBMS as an operating system (or set of subsystems), designed
specifically for controlling data access; its primary functions are storing, retrieving, and securing
data. An RDBMS maintains its own list of authorized users and their associated privileges;
manages memory caches and paging; controls locking for concurrent resource usage; dispatches
and schedules user requests; and manages space usage within its table-space structures
.
72. Name the sub-systems of a RDBMS
I/O, Security, Language Processing, Process Control, Storage Management, Logging and
Recovery, Distribution Control, Transaction Control, Memory Management, Lock Management
73. Which part of the RDBMS takes care of the data dictionary? How
Data dictionary is a set of tables and database objects that is stored in a special area of the
database and maintained exclusively by the kernel.
74. What is the job of the information stored in data-dictionary?
The information in the data dictionary validates the existence of the objects, provides
access to them, and maps the actual physical storage location.
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14. Write a program to demonstrate user defined
exceptions
AIM: to demonstrate user defined exceptions
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Declare
A number
B number;
C number;
Mydivide_zero EXCEPTION;
Begin
A:=&a;
B:=&b;
If(B=0) then
Raise Mydivide_zero;
else
C:=a/b;
Dbms_output.put_line(‘division is ‘ || C);
End if;
Exception
If (mydivide_zero) then
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Dbms_output.put_line(‘b could not be zero’);
End if;
End;
Valid Test Data:
Enter the value for a:
10
Enter the value for b:
0
OUTPUT:
b could not be zero
Viva-Vice:
75. Not only RDBMS takes care of locating data it also
determines an optimal access path to store or retrieve the data
76. How do you communicate with an RDBMS?
You communicate with an RDBMS using Structured Query Language (SQL)
77. Define SQL and state the differences between SQL and other conventional programming
Languages
SQL is a nonprocedural language that is designed specifically for data access operations
on normalized relational database structures. The primary difference between SQL and other
conventional programming languages is that SQL statements specify what data operations should
be performed rather than how to perform them.
78. Name the three major set of files on disk that compose a database in Oracle
There are three major sets of files on disk that compose a database. All the files are binary. These are
 Database files
 Control files
 Redo logs
The most important of these are the database files where the actual data resides. The
control files and the redo logs support the functioning of the architecture itself.
All three sets of files must be present, open, and available to Oracle for any data on the database to be useable. Without these files,
you cannot access the database, and the database administrator might have to recover some or all of the database using a backup, if there is one.
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79. What is an Oracle Instance?
The Oracle system processes, also known as Oracle background processes, provide
functions for the user processes—functions that would otherwise be done by the user processes
themselves
Oracle database-wide system memory is known as the SGA, the system global area or
shared global area. The data and control structures in the SGA are shareable, and all the Oracle
background processes and user processes can use them.
The combination of the SGA and the Oracle background processes is known as an Oracle
instance
80. What are the four Oracle system processes that must always be up and running for the
database to be useable
The four Oracle system processes that must always be up and running for the database to
be useable include DBWR (Database Writer), LGWR (Log Writer), SMON (System Monitor),
and PMON (Process Monitor).
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15. Create a Cursor which update the salaries of an
Employee as follows.
1. if sal<1000then update the salary to 1500.
2. if sal>=1000 and <2000 then update the salary to 2500.
3. if sal>=2000 and <=3000 then update the salary to 4000.
And also count the no.of records have been updated.*/
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Declare
Cursor my_cur is select empno,sal from emp;
Xno emp.empno%type;
Xsal emp.sal%type;
C number;
Begin
Open my_cur;
C:=0;
Loop
Fetch my_cur into xno,xsal;
If(xsal<1000) then
Update emp set sal=3000 where empno=xno;
C:=c+1;
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Else if(xsal>=2000) and xsa<3000) then
Update emp set sal=4000 where empno=xno;
C:=c+1;
End if;
End if;
Exit when my_cur%NOTFOUND
;
End loop;
Close my_cur;
Dbma_output.put_line(c||’records have been successfully updated’);
End;
Sql>@a.sql;
records have been successfully updated
pl/sql procedure successfully completed.
Valid Test Data
Before executing the cursor, the records in emp table as follows
Sql>select * from emp;
OUTPUT:
EMPNO ENAME
JOB MGR HIREDATE SAL COMMD EPTNO
----------------------------------------------------------------7369 SMITH
CLERK
7499 ALLEN
SALESMAN
7521 WARD
SALESMAN
EMPNO ENAME JOB
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7902 17-DEC-80
2000
7698 20-FEB-81
7698 22-FEB-81
MGR HIREDATE
SAL
20
1600
300
30
1250
500
30
COMM
DEPTNO
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-------- ---------- --------- ---------- --------- ---------- --------------------------------------7566 JONES
MANAGER
7654 MARTIN SALESMAN
7698 BLAKE
MANAGER
7839 02-APR-81
7698 28-SEP-81
2975
1250 1400
7839 01-MAY-81
2850
20
30
30
…
….
…
14 rows selected.
Viva-Vice:
81. What are database files, control files and log files. How many of these files should a
database have at least? Why?
Database Files
The database files hold the actual data and are typically the largest in size.
Depending on their sizes, the tables (and other objects) for all the user accounts can go in one
database file—but that's not an ideal situation because it does not make the database structure
very flexible for controlling access to storage for different users, putting the database on different
disk drives, or backing up and restoring just part of the database.
You must have at least one database file but usually, more than one files are used.
In terms of accessing and using the data in the tables and other objects, the number (or location)
of the files is immaterial.
The database files are fixed in size and never grow bigger than the size at which
they were created
Control Files
The control files and redo logs support the rest of the architecture. Any
database must have at least one control file, although you typically have more than one to
guard against loss. The control file records the name of the database, the date and time it was
created, the location of the database and redo logs, and the synchronization information to
ensure that all three sets of files are always in step. Every time you add a new database or
redo log file to the database, the information is recorded in the control files.
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Redo Logs
Any database must have at least two redo logs. These are the journals for the
database; the redo logs record all changes to the user objects or system objects. If any type of
failure occurs, the changes recorded in the redo logs can be used to bring the database to a
consistent state without losing any committed transactions. In the case of non-data loss failure,
Oracle can apply the information in the redo logs automatically without intervention from the
DBA.
The redo log files are fixed in size and never grow dynamically from the size at
which they were created.
82. What is ROWID?
The ROWID is a unique database-wide physical address for every row on every table.
Once assigned (when the row is first inserted into the database), it never changes until the row is
deleted or the table is dropped.
The ROWID consists of the following three components, the combination of which
uniquely identifies the physical storage location of the row.
 Oracle database file number, which contains the block with the rows
 Oracle block address, which contains the row
 The row within the block (because each block can hold many rows)
The ROWID is used internally in indexes as a quick means of retrieving rows with a
particular key value. Application developers also use it in SQL statements as a quick way to
access a row once they know the ROWID
83. What is Oracle Block? Can two Oracle Blocks have the same address?
Oracle "formats" the database files into a number of Oracle blocks when they are first
created—making it easier for the RDBMS software to manage the files and easier to read data
into the memory areas.
The block size should be a multiple of the operating system block size. Regardless of the
block size, the entire block is not available for holding data; Oracle takes up some space to
manage the contents of the block. This block header has a minimum size, but it can grow.
These Oracle blocks are the smallest unit of storage. Increasing the Oracle block size can
improve performance, but it should be done only when the database is first created.
Each Oracle block is numbered sequentially for each database file starting at 1. Two
blocks can have the same block address if they are in different database files.
84. What is database Trigger?
A database trigger is a PL/SQL block that can defined to automatically execute for insert,
update, and delete statements against a table. The trigger can e defined to execute once for the
entire statement or once for every row that is inserted, updated, or deleted. For any one table,
there are twelve events for which you can define database triggers. A database trigger can call
database procedures that are also written in PL/SQL.
85. Name two utilities that Oracle provides, which are use for backup and recovery.
Along with the RDBMS software, Oracle provides two utilities that you can use to back
up and restore the database. These utilities are Export and Import.
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The Export utility dumps the definitions and data for the specified part of the database to
an operating system binary file. The Import utility reads the file produced by an export, recreates
the definitions of objects, and inserts the data
If Export and Import are used as a means of backing up and recovering the database, all
the changes made to the database cannot be recovered since the export was performed. The best
you can do is recover the database to the time when the export was last performed.
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19. create a procedure which generate all the prime
numbers below the given number and count the no.of
prime numbers.
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Create or replace procedure prime_proc(n IN number,tot OUT number)
as
i number;
c number;
j number;
Begin
i:=1;
tot:=0;
while(i<=n)
loop
j:=1;
c:=0;
while(j<=i)
loop
if(mod(I,j)=0) then
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c:=c+1;
end if;
j:=j+1;
end loop;
if(c=2) then
dbms_output.put_line(i);
tot:=tot+1;
end if;
i:=i+1;
end loop;
end;
/
Sql>procedure created.
declare
t number;
begin
prime_proc(10,t);
dbms_output.put_line(‘the total prime no .are’||t);
end;
Valid Test Data:
sql>set serveroutput on
OUTPUT
sql>/
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2
3
5
7
The total prime no.are 4
Pl/sql procedure successfully completed.
Viva-Vice
86. Name two utilities that Oracle provides, which are use for backup and recovery.
Along with the RDBMS software, Oracle provides two utilities that you can use to back
up and restore the database. These utilities are Export and Import.
The Export utility dumps the definitions and data for the specified part of the database to
an operating system binary file. The Import utility reads the file produced by an export, recreates
the definitions of objects, and inserts the data
If Export and Import are used as a means of backing up and recovering the database, all
the changes made to the database cannot be recovered since the export was performed. The best
you can do is recover the database to the time when the export was last performed.
87. What are stored-procedures? And what are the advantages of using them.
Stored procedures are database objects that perform a user defined operation. A stored
procedure can have a set of compound SQL statements. A stored procedure executes the SQL
commands and returns the result to the client. Stored procedures are used to reduce network
traffic.
88. Tables derived from the ERD
a) Are totally unnormalised
b) Are always in 1NF
c) Can be further denormalised
d) May have multi-valued attributes
(b) Are always in 1NF
89. Spurious tuples may occur due to
i. Bad normalization
ii. Theta joins
iii. Updating tables from join
a) i & ii
b) ii & iii
c) i & iii
d) ii & iii
(a) i & iii because theta joins are joins made on keys that are not primary keys.
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90. A B C is a set of attributes. The functional dependency is as follows
AB -> B
AC -> C
C -> B
a) is in 1NF
b) is in 2NF
c) is in 3NF
d) is in BCNF
(a) is in 1NF since (AC)+ = { A, B, C} hence AC is the primary key. Since
C
B
is a FD given, where neither C is a Key nor B is a prime attribute, this it is not in 3NF. Further B
is not functionally dependent on key AC thus it is not in 2NF. Thus the given FDs is in 1NF.
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18. create a procedure which updates the salaries of
an employees as follows.
1.if sal<1000 then update the salry to 1500.
2.if sal>=1000 and <=2400 then update the salary to 2500.*/
HW/SW requirements:
Processor
:
AMD Athelon ™ 1.67 GHz
RAM
:
256 MB
Hard Disk
:
40 GB
Software
:
Oracle, PlSQL
Create or replace procedure myproc as
Cursor my_cur is select empno,sal from emp;
Xno emp.empno%type;
Xsal emp.sal%type;
C number;
Begin
Open my_cur;
C:=0;
Loop
Fetch my_cur into xno,xsal;
If(xsal<1000) then
Update emp set sal=1500 where empno=xno;
C:=c+1;
Else
Is(xsal>=1000 and xsal<=2400) then
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Update emp set sal=2500 where empno=xno;
C:=c+1;
End if;
End if;
Exit when my_cur%NOTFOUND;
End loop;
Close my_cur;
Dbms_output.put_line(c||’records have been successfully updated’);
End;
/
Valid Test Data:
Procedure created.
Sql>exec myproc;
OUTPUT:
Records have been successfully completed.
/* create function which add two given numbers. (Simple programs) */
Create or replace function add_fun(a number,b number) return
Number as
C number;
Begin
C:=a+b;
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Return c;
End;
/
Function created.
/*add_fun specification*/
Declare
Result number;
Begin
Result:=add_fun(10,20);
Dbms_output.put_line(‘the sum of 10 and 20 is’||result);
End;
Sql>/
The sum of 10 and 20 is 30
Pl/sql procedure successfully completed.
/*create a function which count total no.of employees having salary
less than 6000.*/
/*function body*/
Create or replace function count_emp(esal number)return number as
Cursor vin_cur as Select empno,sal from emp;
Xno emp.empno%type;
Xsal emp.sal%type;
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C number;
Begin
Open vin_cur;
C:=0;
loop
fetch vin_cur into xno,xsal;
if(xsal<esal) then
c:=c+1;
end if;
exit when vin_cur%notfound;
end loop;
close vin_cur;
return c;
end;
/
Function created.
/*function specification*/
Declare
Ne number;
Xsal number;
Begin
Ne:=count_emp(xsal);
Dbms_output.put_line(xsal);
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Dbma_output.put_line(‘there are ‘||ne||;employees’);
End;
/
OUTPUT
There are 8 employees.
Viva-Vice:
91. In mapping of ERD to DFD
a) entities in ERD should correspond to an existing entity/store in DFD
b) entity in DFD is converted to attributes of an entity in ERD
c) relations in ERD has 1 to 1 correspondence to processes in DFD
d) relationships in ERD has 1 to 1 correspondence to flows in DFD
(a) entities in ERD should correspond to an existing entity/store in DFD
92. A dominant entity is the entity
a) on the N side in a 1 : N relationship
b) on the 1 side in a 1 : N relationship
c) on either side in a 1 : 1 relationship
d) nothing to do with 1 : 1 or 1 : N relationship
(b) on the 1 side in a 1 : N relationship
93. Select 'NORTH', CUSTOMER From CUST_DTLS Where REGION = 'N' Order By
CUSTOMER Union Select 'EAST', CUSTOMER From CUST_DTLS Where REGION = 'E'
Order By CUSTOMER
The above is
a) Not an error
b) Error - the string in single quotes 'NORTH' and 'SOUTH'
c) Error - the string should be in double quotes
d) Error - ORDER BY clause
(d) Error - the ORDER BY clause. Since ORDER BY clause cannot be used in UNIONS
94. What is Storage Manager?
It is a program module that provides the interface between the low-level data stored in
database, application programs and queries submitted to the system.
95. What is Buffer Manager?
It is a program module, which is responsible for fetching data from disk storage into main
memory and deciding what data to be cache in memory.
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96. What is Transaction Manager?
It is a program module, which ensures that database, remains in a consistent state despite
system failures and concurrent transaction execution proceeds without conflicting.
97. What is File Manager?
It is a program module, which manages the allocation of space on disk storage and data
structure used to represent information stored on a disk.
98. What is Authorization and Integrity manager?
It is the program module, which tests for the satisfaction of integrity constraint and
checks the authority of user to access data.
99. What are stand-alone procedures?
Procedures that are not part of a package are known as stand-alone because they
independently defined. A good example of a stand-alone procedure is one written in a
SQL*Forms application. These types of procedures are not available for reference from other
Oracle tools. Another limitation of stand-alone procedures is that they are compiled at run time,
which slows execution.
100.
What are cursors give different types of cursors.
PL/SQL uses cursors for all database information accesses statements. The language
supports the use two types of cursors
 Implicit
 Explicit
101. What is cold backup and hot backup (in case of Oracle)?
 Cold Backup:
It is copying the three sets of files (database files, redo logs, and control file)
when the instance is shut down. This is a straight file copy, usually from the disk directly to tape.
You must shut down the instance to guarantee a consistent copy.
If a cold backup is performed, the only option available in the event of data
file loss is restoring all the files from the latest backup. All work performed on the database since
the last backup is lost.
 Hot Backup:
Some sites (such as worldwide airline reservations systems) cannot shut down
the database while making a backup copy of the files. The cold backup is not an available option.
So different means of backing up database must be used — the hot backup.
Issue a SQL command to indicate to Oracle, on a tablespace-by-tablespace basis, that the files of
the tablespace are to backed up. The users can continue to make full use of the files, including
making changes to the data. Once the user has indicated that he/she wants to back up the
tablespace files, he/she can use the operating system to copy those files to the desired backup
destination.
The database must be running in ARCHIVELOG mode for the hot backup
option.
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If a data loss failure does occur, the lost database files can be restored using
the hot backup and the online and offline redo logs created since the backup was done. The
database is restored to the most consistent state without any loss of committed transactions.
102.
What are Armstrong rules? How do we say that they are complete and/or sound
The well-known inference rules for FDs
 Reflexive rule :
If Y is subset or equal to X then X
Y.
 Augmentation rule:
If X
Y then XZ
YZ.
 Transitive rule:
If {X
Y, Y
Z} then X
Z.
 Decomposition rule :
If X
YZ then X
Y.
 Union or Additive rule:
If {X
Y, X
Z} then X
YZ.
 Pseudo Transitive rule :
If {X
Y, WY
Z} then WX
Z.
Of these the first three are known as Amstrong Rules. They are sound because it is
enough if a set of FDs satisfy these three. They are called complete because using these three
rules we can generate the rest all inference rules.
103.
How can you find the minimal key of relational schema?
Minimal key is one which can identify each tuple of the given relation schema uniquely.
For finding the minimal key it is required to find the closure that is the set of all attributes that
are dependent on any given set of attributes under the given set of functional dependency.
Algo. I Determining X+, closure for X, given set of FDs F
1. Set X+ = X
2. Set Old X+ = X+
3. For each FD Y
Z in F and if Y belongs to X+ then add Z to X+
4. Repeat steps 2 and 3 until Old X+ = X+
Algo.II Determining minimal K for relation schema R, given set of FDs F
1. Set K to R that is make K a set of all attributes in R
2. For each attribute A in K
a. Compute (K – A)+ with respect to F
b. If (K – A)+ = R then set K = (K – A)+
104.
What do you understand by dependency preservation?
Given a relation R and a set of FDs F, dependency preservation states that the
closure of the union of the projection of F on each decomposed relation Ri is equal to the
closure of F. i.e.,
((R1(F)) U … U (Rn(F)))+ = F+
if decomposition is not dependency preserving, then some dependency is lost in the
decomposition.
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105.
What is meant by Proactive, Retroactive and Simultaneous Update.
Proactive Update:
The updates that are applied to database before it becomes effective in real
world .
Retroactive Update:
The updates that are applied to database after it becomes effective in real
world .
Simulatneous Update:
The updates that are applied to database at the same time when it becomes
effective in real world .
106.
What are the different types of JOIN operations?
Equi Join: This is the most common type of join which involves only equality
comparisions. The disadvantage in this type of join is that there
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SQL Questions:
1. Which is the subset of SQL commands used to manipulate Oracle Database structures,
including tables?
Data Definition Language (DDL)
2. What operator performs pattern matching?
LIKE operator
3. What operator tests column for the absence of data?
IS NULL operator
4. Which command executes the contents of a specified file?
START <filename> or @<filename>
5. What is the parameter substitution symbol used with INSERT INTO command?
&
6. Which command displays the SQL command in the SQL buffer, and then executes it?
RUN
7. What are the wildcards used for pattern matching?
_ for single character substitution and % for multi-character substitution
8. State true or false. EXISTS, SOME, ANY are operators in SQL.
True
9. State true or false. !=, <>, ^= all denote the same operation.
True
10. What are the privileges that can be granted on a table by a user to others?
Insert, update, delete, select, references, index, execute, alter, all
11. What command is used to get back the privileges offered by the GRANT command?
REVOKE
12. Which system tables contain information on privileges granted and privileges obtained?
USER_TAB_PRIVS_MADE, USER_TAB_PRIVS_RECD
13. Which system table contains information on constraints on all the tables created?
USER_CONSTRAINTS
14.
TRUNCATE TABLE EMP;
DELETE FROM EMP;
Will the outputs of the above two commands differ?
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Both will result in deleting all the rows in the table EMP.
15. What is the difference between TRUNCATE and DELETE commands?
TRUNCATE is a DDL command whereas DELETE is a DML command. Hence
DELETE operation can be rolled back, but TRUNCATE operation cannot be rolled back.
WHERE clause can be used with DELETE and not with TRUNCATE.
16. What command is used to create a table by copying the structure of another table?
Answer :
CREATE TABLE .. AS SELECT command
Explanation :
To copy only the structure, the WHERE clause of the SELECT command should contain
a FALSE statement as in the following.
CREATE TABLE NEWTABLE AS SELECT * FROM EXISTINGTABLE WHERE
1=2;
If the WHERE condition is true, then all the rows or rows satisfying the condition will be
copied to the new table.
17. What will be the output of the following query?
SELECT REPLACE(TRANSLATE(LTRIM(RTRIM('!! ATHEN !!','!'), '!'), 'AN',
'**'),'*','TROUBLE') FROM DUAL;
TROUBLETHETROUBLE
18. What will be the output of the following query?
SELECT DECODE(TRANSLATE('A','1234567890','1111111111'), '1','YES', 'NO' );
Answer :
NO
Explanation :
The query checks whether a given string is a numerical digit.
19. What does the following query do?
SELECT SAL + NVL(COMM,0) FROM EMP;
This displays the total salary of all employees. The null values in the commission
column will be replaced by 0 and added to salary.
20. Which date function is used to find the difference between two dates?
MONTHS_BETWEEN
21. Why does the following command give a compilation error?
DROP TABLE &TABLE_NAME;
Variable names should start with an alphabet. Here the table name starts with an '&'
symbol.
22. What is the advantage of specifying WITH GRANT OPTION in the GRANT command?
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The privilege receiver can further grant the privileges he/she has obtained from the
owner to any other user.
23. What is the use of the DROP option in the ALTER TABLE command?
It is used to drop constraints specified on the table.
24. What is the value of ‘comm’ and ‘sal’ after executing the following query if the initial value
of ‘sal’ is 10000?
UPDATE EMP SET SAL = SAL + 1000, COMM = SAL*0.1;
sal = 11000, comm = 1000
25. What is the use of DESC in SQL?
Answer :
DESC has two purposes. It is used to describe a schema as well as to retrieve rows from
table in descending order.
Explanation :
The query SELECT * FROM EMP ORDER BY ENAME DESC will display the output
sorted on ENAME in descending order.
26. What is the use of CASCADE CONSTRAINTS?
When this clause is used with the DROP command, a parent table can be dropped even
when a child table exists.
27. Which function is used to find the largest integer less than or equal to a specific value?
FLOOR
28. What is the output of the following query?
SELECT TRUNC(1234.5678,-2) FROM DUAL;
1200
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REFERENCES:
SCOTT Urman, oracle 8i-PL/SQL programming,TMH-2000
Loney, oracle 8i-the complete reference,TMH-2000
Loney, oracle 9i-the complete reference,TMH-2000
Bayross, oracle Teach Your Self SQL/PLSQL using oracle 8i and 9i
with SQLJ, BPB, 2002.
 Abbey, oracle 8i-A beginners guide,TMH-2000.




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Annexure - 1
GURU NANAK Engineering College
Ibrahimpatnam, R R District – 501 506 (A. P.)
Assignment / Record No - _____
Department of ________________________
Name
:
______________________________________
Class / Year /
Semester
:
B. Tech _____ Year _________ Semester
Roll No.
:
_________________________
Subject
:
_________________________
Branch
:
_________________________
Date of
Submission
:
_____ / _____ / __________
________________
Student Signature
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Faculty Signature
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Annexure - 2
Index Page
S. No.
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Date
Program / Experiment
Page
No.
Sign /
Remarks
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Annexure - 3
DBMS Lab Manual
GURU NANAK Engineering College
Ibrahimpatnam, R R District – 501 506 (A. P.)
(Sponsored by: Guru Nanak Educational Society, Hyderabad (A. P.)
Department of ________________________
This is to certify that
Name
:
______________________________________
Class / Year /
Semester
:
B. Tech ______ Year ________ Semester
Roll No.
:
_________________________
Subject
:
_________________________
Branch
:
_________________________
Has successfully completed the course of programs /
experiments prescribed by Jawharlal Nerhu
Technological Univeristy of the department during the
academic year 200__ - 200 __.
________________
Faculty Incharge
jaipal
________________
Department Head
97