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Revision for Mid 1
Functional Dependencies
 FDs defined over two sets of
attributes: X,
R
YR
 Notation: X  Y reads as “X
determines Y”
 If X  Y, then all tuples that agree
X
Y
Z
1
2
3
2
4
5
1
2
4
1
2
7
2
4
8
3
7
9
on X must also agree on Y
Database System Concepts
7.2
©Silberschatz, Korth and Sudarshan
Functional Dependencies (example)
X
Database System Concepts
Y
Z
7.3
X
Y
Z
1
2
3
2
4
5
1
2
4
1
2
7
2
4
8
3
7
9
©Silberschatz, Korth and Sudarshan
Candidate Keys


an attribute (or set of attributes) that uniquely identifies a row
primary key is a special candidate key


values cannot be null
e.g.

ENROLL (Student_ID, Name, Address, …)
 PK = Student_ID
 candidate key = Name, Address
Database System Concepts
7.4
©Silberschatz, Korth and Sudarshan
Database System Concepts
7.5
©Silberschatz, Korth and Sudarshan
2NF
 a relation is in second normal form if it is in first normal form
AND every nonkey attribute is fully functionally dependant on the
primary key
 i.e. remove partial functional dependencies, so no nonkey
attribute depends on just part of the key
Database System Concepts
7.6
©Silberschatz, Korth and Sudarshan
EMPLOYEE2 (Emp_ID, Course_Title, Name,
Dept_Name, Salary, Date_Completed)
Emp_ID
Course_
Title
Name
Dept_
Name
Salary
Date_Com
p.
not fully functionally
dependant on the primary key
Database System Concepts
7.7
©Silberschatz, Korth and Sudarshan
Second Normal Form ( 2NF )
 it is based on the concept of full functional dependency.
 A functional dependency XY is a full functional
dependency , for any attribute A  X, {X - {A}}  Y.
Database System Concepts
7.8
©Silberschatz, Korth and Sudarshan
2NF
A
B
(Example)
C
D
R (A B C D)
1
1
2
3
2
1
3
2
2 Candidate Keys
3
1
2
3
R with key{AB} is NOT 2NF
1
2
1
3
R with key{AC} is NOT 2NF
Database System Concepts
7.9
©Silberschatz, Korth and Sudarshan
Second Normal Form
Second normal form:
Let R’ be a relation, and let F be the set of governing FDs. An attribute
belongs to R’ is prime if a key of R’ contains A. In other words, A is prime in R’
if there exists K<R’ such that (1) K->R’,
(2) for all B belongs to K, (K-B)->R’ not belongs to
F+, and
(3) A belongs to K
Database System Concepts
7.10
©Silberschatz, Korth and Sudarshan
Database System Concepts
7.11
©Silberschatz, Korth and Sudarshan
General Definitions of Second Normal Form
 A relation schema R is in second normal form (2NF) if
every nonprime attribute A in R is fully functionally
dependent on every key of R.
Database System Concepts
7.12
©Silberschatz, Korth and Sudarshan
Third Normal Form
A relation R is in 3NF if,
for all X  A that holds over R

A  X ( i.e., X  A is a trivial FD ), or

X is a superkey, or

A is part of some key for R
If R is in BCNF,
obviously it is in
3NF.

The definition of 3NF is similar to that of BCNF, with the only difference being the third
condition.

Recall that a key for a relation is a minimal set of attributes that uniquely determines all other
attributes.

A must be part of a key (any key, if there are several).

It is not enough for A to be part of a superkey, because this condition is satisfied by every attribute.
Database System Concepts
7.13
©Silberschatz, Korth and Sudarshan
 Suppose that a dependency X  A causes a violation of 3NF.
There are two cases:
 X is a proper subset of some key K. Such a dependency is
sometimes called a partial dependency. In this case, we store
(X,A) pairs redundantly.
 X is not a proper subset of any key. Such a dependency is
sometimes called a transitive dependency, because it means we
have a chain of dependencies K  XA.
Database System Concepts
7.14
©Silberschatz, Korth and Sudarshan
Key
Attributes X
Attributes A
A not in a key
Partial Dependencies
Key
Key
Attributes X
Attributes A
Attributes A
Attributes X
A not in a key
A in a key
Transitive Dependencies
Database System Concepts
7.15
©Silberschatz, Korth and Sudarshan
 Motivation of 3NF
 By making an exception for certain dependencies involving key
attributes, we can ensure that every relation schema can be
decomposed into a collection of 3NF relations using only
decompositions.
 Such a guarantee does not exist for BCNF relations.
 It weaken the BCNF requirements just enough to make this
guarantee possible.
 Unlike BCNF, some redundancy is possible with 3NF.
 The problems associate with partial and transitive dependencies
persist if there is a nontrivial dependency XA and X is not a
superkey, even if the relation is in 3NF because A is part of a key.
Database System Concepts
7.16
©Silberschatz, Korth and Sudarshan
Reserves

Assume: sid  cardno (a sailor uses a unique credit card to pay for reservations).

Reserves is not in 3NF



sid is not a key and cardno is not part of a key
In fact, (sid, bid, day) is the only key.
(sid, cardno) pairs are redundantly.
Database System Concepts
7.17
©Silberschatz, Korth and Sudarshan
Reserves

Assume: sid  cardno, and cardno  sid (we know that credit cards also uniquely
identify the owner).

Reserves is in 3NF


(cardno, sid, bid) is also a key for Reserves.
sid  cardno does not violate 3NF.
Database System Concepts
7.18
©Silberschatz, Korth and Sudarshan
Lecture 12:
Further relational algebra,
further SQL
www.cl.cam.ac.uk/Teaching/current/Databases/
Today’s lecture
 Where does SQL differ from relational model?
 What are some other features of SQL?
 How can we extend the relational algebra to match more closely
SQL?
Database System Concepts
7.20
©Silberschatz, Korth and Sudarshan
Duplicate rows
 Consider our relation instances from lecture 6, Reserves, Sailors
and Boats
 Consider
SELECT rating,age
FROM
Sailors;
 We get a relation that doesn’t satisfy our definition of a relation!
 RECALL: We have the keyword DISTINCT to remove duplicates
Database System Concepts
7.21
©Silberschatz, Korth and Sudarshan
Multiset semantics
 A relation in SQL is really a multiset or bag, rather than a set as
in the relational model
 A multiset has no order (unlike a list), but allows duplicates
 E.g. {1,2,1,3} is a bag
 select, project and join work for bags as well as sets
 Just work on a tuple-by-tuple basis
Database System Concepts
7.22
©Silberschatz, Korth and Sudarshan
Extended relational algebra
Add features needed for SQL
1.
Bag semantics
2.
Duplicate elimination operator, 
3.
Sorting operator, 
4.
Grouping and aggregation operator, 
Outerjoin operators, oV, Vo, oVo
5.
Database System Concepts
7.23
©Silberschatz, Korth and Sudarshan
Duplicate-elimination operator
 (R) = relation R with any duplicated tuples removed
(R)=
 R=
A
B
A
B
 This is used
1
2to model the DISTINCT
1
2 feature of SQL
Database System Concepts
3
4
1
2
3
4
7.24
©Silberschatz, Korth and Sudarshan
Sorting

L1,… Ln(R) returns a list of tuples of R, ordered according to the attributes L1, …, Ln

Note:  does not return a relation

R=

B(R)= [(5,2),(1,3),(3,4)]
A
B
1
3
3
4
ORDER BY in SQL, e.g.
SELECT *
5
2
FROM Sailors
WHERE rating>7
ORDER BY age, sname;
Database System Concepts
7.25
©Silberschatz, Korth and Sudarshan
Extended projection
 SQL allows us to use arithmetic operators
SELECT age*5
FROM
Sailors;
 We extend the projection operator to allow the columns in the projection to be
functions of one or more columns in the argument relation, e.g.
A+B,A,A(R)=
 R=
Database System Concepts
A
B
A+B
A.1
A.2
1
2
3
1
1
3
4
7
3
3
7.26
©Silberschatz, Korth and Sudarshan
Arithmetic
 Arithmetic (and other expressions) can not be used at the top
level
 i.e. 2+2 is not a valid SQL query
 How would you get SQL to compute 2+2?
Database System Concepts
7.27
©Silberschatz, Korth and Sudarshan
Aggregation
 SQL provides us with operations to summarise a column in some way,
e.g.
SELECT COUNT(rating)
FROM
Sailors;
SELECT COUNT(DISTINCT rating)
FROM
Sailors;
SELECT COUNT(*)
FROM
Sailors WHERE rating>7;
 We also have SUM, AVG, MIN and MAX
Database System Concepts
7.28
©Silberschatz, Korth and Sudarshan
Grouping
 These aggregation operators have been applied to all qualifying
tuples. Sometimes we want to apply them to each of several
groups of tuples, e.g.
 For each rating, find the average age of the sailors
 For each rating, find the age of the youngest sailor
Database System Concepts
7.29
©Silberschatz, Korth and Sudarshan
GROUP BY in SQL
SELECT [DISTINCT] target-list
FROM
relation-list
WHERE
qualification
GROUP BY grouping-list;

The target-list contains
1.
List of column names
2.
Aggregate terms

NOTE: The variables in target-list must be contained in groupinglist
Database System Concepts
7.30
©Silberschatz, Korth and Sudarshan
GROUP BY cont.
For each rating, find the average age of the sailors
SELECT
rating,AVG(age)
FROM
Sailors
GROUP BY rating;
For each rating find the age of the youngest sailor
SELECT
rating,MIN(age)
FROM
Sailors
GROUP BY rating;
Database System Concepts
7.31
©Silberschatz, Korth and Sudarshan
Grouping and aggregation
 L(R) where L is a list of elements that are either
 Individual column names (“Grouping attributes”), or
 Of the form (A), where  is an aggregation operator (MIN, SUM, …)
and A is the column it is applied to
 For example,
rating,AVG(age)(Sailors)
Database System Concepts
7.32
©Silberschatz, Korth and Sudarshan
Example
 Let R=
bar
beer
price
Anchor
6X
2.50
Anchor
Adnam’s
2.40
Mill
6X
2.60
Mill
Fosters
2.80
Eagle
Fosters
2.90
 Compute beer,AVG(price)(R)
Database System Concepts
7.33
©Silberschatz, Korth and Sudarshan
Example cont.
1.
2.
Group according to the grouping attribute, beer:
bar
beer
price
Anchor
6X
2.50
Mill
6X
2.60
Anchor
Adnam’s
2.40
Fosters
2.80
Fosters
2.90
Compute average
Mill
Eagle
Database System Concepts
7.34
of price within groups:
beer
price
6X
2.55
Adnam’s
2.40
Fosters
2.85
©Silberschatz, Korth and Sudarshan
NULL values
 Sometimes field values are unknown (e.g. rating not known yet), or
inapplicable (e.g. no spouse name)
 SQL provides a special value, NULL, for both these situations
 This complicates several issues
 Special operators needed to check for NULL
 Is NULL>8? Is (NULL OR TRUE)=TRUE?
 We need a three-valued logic
 Need to carefully re-define semantics
Database System Concepts
7.35
©Silberschatz, Korth and Sudarshan
NULL values
 Consider
INSERT INTO Sailors (sid,sname)
VALUES (101,”Julia”);
SELECT * FROM Sailors;
SELECT rating FROM Sailors;
SELECT sname
FROM
Sailors
WHERE
rating>0;
Database System Concepts
7.36
©Silberschatz, Korth and Sudarshan
Entity integrity constraint
 An entity integrity constraint states that no primary key value
can be NULL
Database System Concepts
7.37
©Silberschatz, Korth and Sudarshan
Outer join
 Note that with the usual join, a tuple that doesn’t ‘join’ with any
from the other relation is removed from the resulting relation
 Instead, we can ‘pad out’ the columns with NULLs
 This operator is called an full outer join, written oVo
Database System Concepts
7.38
©Silberschatz, Korth and Sudarshan
Example of full outer join
 Let R=
Let S=
A
B
B
C
1
2
4
5
 Then RVS =3
4
6
7
 But RoVoS =
Database System Concepts
A
B
C
3
4
5
A
B
C
1
2
NULL
3
4
5
NULL
6
7
7.39
©Silberschatz, Korth and Sudarshan
Outer joins in SQL
 SQL/92 has three variants:
 LEFT OUTER JOIN (algebra: oV)
 RIGHT OUTER JOIN (algebra: Vo)
 FULL OUTER JOIN
(algebra: oVo)
 For example:
SELECT *
FROM
Reserves r LEFT OUTER JOIN
Sailors s
ON
r.sid=s.sid;
Database System Concepts
7.40
©Silberschatz, Korth and Sudarshan
Views
 A view is a query with a name that can be used in further SELECT
statements, e.g.
CREATE VIEW
ExpertSailors(sid,sname,age)
AS SELECT sid,sname,age
FROM
Sailors
WHERE
rating>9;
 Note that ExpertSailors is not a stored relation
 (WARNING: mysql does not support views )
Database System Concepts
7.41
©Silberschatz, Korth and Sudarshan
Querying views
 So an example query
SELECT sname
FROM
ExpertSailors
WHERE
age>27;
 is translated by the system to the following:
SELECT sname
FROM
Sailors
WHERE
rating>9 AND age>27;
Database System Concepts
7.42
©Silberschatz, Korth and Sudarshan
Relational Algebra
 The Relational Algebra is used to define the ways in which relations
(tables) can be operated to manipulate their data.
 It is used as the basis of SQL for relational databases, and
illustrates the basic operations required of any DML.
 This Algebra is composed of Unary operations (involving a single
table) and Binary operations (involving multiple tables).
Database System Concepts
7.43
©Silberschatz, Korth and Sudarshan
SQL
 Structured Query Language (SQL)
 Standardised by ANSI
 Supported by modern RDBMSs
 Commands fall into three groups

Data Definition Language (DLL)
 Create tables, etc
 Data Manipulation Language (DML)
 Retrieve and modify data
 Data Control Language
 Control what users can do – grant and revoke privileges
Database System Concepts
7.44
©Silberschatz, Korth and Sudarshan
Selection
 The selection or  operation selects rows from a table that satisfy a condition:
 < condition > < tablename >
 Example:
 course = ‘CM’ Students
Students
stud#
name
course
100
Fred
PH
stud#
name
course
200
Dave
CM
200
Dave
CM
300
Bob
CM
300
Bob
CM
Database System Concepts
7.45
©Silberschatz, Korth and Sudarshan
Projection

The projection or  operation selects a list of columns from a table.
 < column list > < tablename >

Example:  stud#, name Students
Students
stud# name
course
stud#
name
100
PH
100
Fred
Fred
200
Dave
CM
200
300
Bob
CM
300
Database System Concepts
7.46
Dave
Bob
©Silberschatz, Korth and Sudarshan
Selection / Projection
 Selection and Projection are usually combined:
 stud#, name ( course = ‘CM’ Students)
Students
stud#
name
course
100Fred PH
stud#
name
200Dave CM
200
Dave
300Bob
300
Bob
CM
Database System Concepts
7.47
©Silberschatz, Korth and Sudarshan
Cartesian Product
 Concatenation of every row in the first relation (R) with every row in
the second relation (S):
R X S
Database System Concepts
7.48
©Silberschatz, Korth and Sudarshan
Cartesian Product - Example
Students
stud# name
100
Fred
200
Dave
300
Bob
Courses
course#
PH
CM
course
PH
CM
CM
name
Pharmacy
Computing
Students X Courses =
stud# Students.name course
100
Fred
100
Fred
200
Dave
200
Dave
300
Bob
300
Bob
Database System Concepts
course#
PH
PH
CM
CM
CM
CM
PH
CM
PH
CM
PH
CM
7.49
Courses.name
Pharmacy
Computing
Pharmacy
Computing
Pharmacy
Computing
©Silberschatz, Korth and Sudarshan
Theta Join
 A Cartesian product with a condition applied:
R ⋈ <condition> S
Database System Concepts
7.50
©Silberschatz, Korth and Sudarshan
Theta Join - Example
Students
Courses
stud# name
course
course# name
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
300
Bob
CM
Students ⋈
stud# = 200
Courses
stud# Students.name course
course#
200
Dave
CM
PH
Pharmacy
200
Dave
CM
CM
Computing
Database System Concepts
Courses.name
7.51
©Silberschatz, Korth and Sudarshan
Inner Join (Equijoin)
 A Theta join where the <condition> is the match (=) of the
primary and foreign keys.
R⋈
Database System Concepts
<R.primary_key = S.foreign_key>
7.52
S
©Silberschatz, Korth and Sudarshan
Inner Join - Example
Students
Courses
stud# name
course
course# name
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
300
Bob
CM
Students ⋈
course = course#
Courses
stud# Students.name course
course#
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
300
Bob
CM
CM
Computing
Database System Concepts
Courses.name
7.53
©Silberschatz, Korth and Sudarshan
Natural Join
 Inner join produces redundant data (in the previous example: course and
course#). To get rid of this duplication:

< stud#, Students.name, course, Courses.name >
(Students ⋈
Or
R1= Students ⋈
R2= 
<course = course#>
Courses)
<course = course#>
Courses
< stud#, Students.name, course, Courses.name >
R1
The result is called the natural join of Students and Courses
Database System Concepts
7.54
©Silberschatz, Korth and Sudarshan
Natural Join - Example
Students
Courses
stud# name
course
course# name
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
300
Bob
R1= Students ⋈
CM
<course = course#>
Courses
R2=  < stud#, Students.name, course, Courses.name > R1
stud# Students.name course
Courses.name
100
Fred
PH
Pharmacy
200
Dave
CM
Computing
300
Bob
CM
Computing
Database System Concepts
7.55
©Silberschatz, Korth and Sudarshan
Outer Joins
 Inner join + rows of one table which do not satisfy the <condition>.
 Left Outer Join: R
<R.primary_key = S.foreign_key>
S
All rows from R are retained and unmatched rows of S are
padded with NULL
 Right Outer Join: R
<R.primary_key = S.foreign_key>
S
All rows from S are retained and unmatched rows of R are
padded with NULL
Database System Concepts
7.56
©Silberschatz, Korth and Sudarshan
Left Outer Join - Example
Students
Courses
stud# name
course
course# name
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
EN
CH
Chemistry
course#
Courses.name
400
Peter
Students
<course = course#>
Courses
stud# Students.name course
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
EN
NULL
NULL
400
Peter
Database System Concepts
7.57
©Silberschatz, Korth and Sudarshan
Right Outer Join - Example
Students
Courses
stud# name
course
course# name
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
EN
CH
Chemistry
400
Peter
Students
<course = course#>
Courses
stud# Students.name course
course#
100
Fred
PH
PH
Pharmacy
200
Dave
CM
CM
Computing
NULL NULL
Database System Concepts
NULL
Courses.name
CH
Chemistry
7.58
©Silberschatz, Korth and Sudarshan
Combination of Unary and Join Operations
Students
stud# name
100
Fred
200
Dave
300
Bob
address
Aberdeen PH
Dundee
Aberdeen CM
course
PH
CM
Courses
course#
name
Pharmacy
CM
Computing
Show the names of students (from Aberdeen) and the names of their courses
R1= Students ⋈ <course=course#> Courses
R2=  <address=“Aberdeen”> R1
R3=  <Students.name, Course.name> R2
Students.name
Fred
Bob
Database System Concepts
7.59
Courses.name
Pharmacy
Computing
©Silberschatz, Korth and Sudarshan
Union
 Takes the set of rows in each table and combines them, eliminating
duplicates
 Participating relations must be compatible, ie have the same number of
columns, and the same column names, domains, and data types
R
S
RS
A
a1
a2
Database System Concepts
B
b1
b2
A
a2
a3
B
b2
b3
7.60
A
a1
a2
a3
B
b1
b2
b3
©Silberschatz, Korth and Sudarshan
Intersection
 Takes the set of rows that are common to each relation
 Participating relations must be compatible
R
A
a1
a2
Database System Concepts
RS
S
B
b1
b2
A
a2
a3
B
b2
b3
7.61
A
a2
B
b2
©Silberschatz, Korth and Sudarshan
Difference
 Takes the set of rows in the first relation but not the second
 Participating relations must be compatible
R
A
a1
a2
Database System Concepts
S
B
b1
b2
A
a2
a3
R-S
B
b2
b3
7.62
A
a1
B
b1
©Silberschatz, Korth and Sudarshan
Exercise (May 2004 Exam)
Employee
WorkLoad
Project
empid
name
empid*
projid*
duration
E100
Fred
E100
P001
17
P001
DB
E200
Dave
E200
P001
12
P002
Access
E300
Bob
E300
P002
15
P003
SQL
E400
Peter
projid
name
Determine the outcome of the following operations:

A natural join between Employee and WorkLoad

A left outer join between Employee and WorkLoad

A right outer join between WorkLoad and Project
Database System Concepts
7.63
©Silberschatz, Korth and Sudarshan
Unary Operations
Selection
Projection
 course = ‘Computing’ Students
 stud#, name Students
In SQL:
In SQL:
Select *
Select stud#, name
From Students
From Students;
Where course = ‘Computing’;
Selection & Projection
 stud#, name ( course = ‘Computing’ Students)
In SQL:
Select stud#, name
From students
Where course = ‘Computing’;
Database System Concepts
7.64
©Silberschatz, Korth and Sudarshan
Binary Operations/Joins
Cartesian Product: Students X Courses
In SQL:
Select *
From Students, Courses;
Theta Join: Students ⋈ <stud# =200> Courses
In SQL:
Select *
From Students, Courses
Where stud# = 200;
Database System Concepts
7.65
©Silberschatz, Korth and Sudarshan
Binary Operations/Joins
Inner Join (Equijoin): Students ⋈ <course=course#> Courses
In SQL:
Select *
From Students, Courses
Where course=course#;
Natural Join:
R1= Students ⋈
<course = course#>
Courses
R2=  < stud#, Students.name, course, Courses.name > R1
In SQL:
Select stud#, Students.name, course, Courses.name
From Students, Courses
Where course=course#;
Database System Concepts
7.66
©Silberschatz, Korth and Sudarshan
Outer Joins
Left Outer Join
Students
<course = course#>
Courses
In SQL:
Select *
From Students, Courses
Where course = course#(+)
Right Outer Join
Students
<course = course#>
Courses
In SQL:
Select *
From Students, Courses
Where course(+) = course#
Database System Concepts
7.67
©Silberschatz, Korth and Sudarshan
Combination of Unary and Join Operations
R1= Students ⋈ <course=course#> Courses
R2=  <address=“Aberdeen”> R1
R3=  <Students.name, Course.name> R2
In SQL:
Select Students.name, Courses.name
From Students, Courses
Where course=course#
AND address=“Aberdeen”;
Database System Concepts
7.68
©Silberschatz, Korth and Sudarshan
Set Operations
Union: R  S
Intersection: R  S
In SQL:
In SQL:
Select * From R
Select * From R
Union
Intersect
Select * From S;
Select * From S;
Difference: R - S
In SQL:
Select * From R
Minus
Select * From S;
Database System Concepts
7.69
©Silberschatz, Korth and Sudarshan
SQL Operators
SELECT *
FROM Book
WHERE catno BETWEEN 200 AND 400;
SELECT *
FROM Product
WHERE prod_desc BETWEEN ‘C’ AND ‘S’;
SELECT *
FROM Book
WHERE catno NOT BETWEEN 200 AND 400;
Database System Concepts
7.70
©Silberschatz, Korth and Sudarshan
SELECT Catno
SQL Operators
FROM Loan
WHERE Date-Returned IS NULL;
SELECT Catno
FROM Loan
WHERE Date-Returned IS NOT NULL;
Database System Concepts
7.71
©Silberschatz, Korth and Sudarshan
SELECT Name
SQL Operators
FROM Member
WHERE memno IN (100, 200, 300, 400);
SELECT Name
FROM Member
WHERE memno NOT IN (100, 200, 300, 400);
Database System Concepts
7.72
©Silberschatz, Korth and Sudarshan
SQL Operators
SELECT Name
FROM Member
WHERE address NOT LIKE ‘%Aberdeen%’;
SELECT Name
FROM Member
WHERE Name LIKE ‘_ES%’;
Note: In MS Access, use * and # instead of % and _
Database System Concepts
7.73
©Silberschatz, Korth and Sudarshan
Selecting Distinct Values
Student
stud#
100 Fred
200 Dave
300 Bob
name
Aberdeen
Dundee
Aberdeen
address
SELECT Distinct address
FROM Student;
address
Aberdeen
Dundee
Database System Concepts
7.74
©Silberschatz, Korth and Sudarshan