Download CSc-340 01b

The Relational Model
Chapter 2
Database Schema
Keys
Schema Diagrams
Relational Query Languages
Relational Operations
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Go Over Homework/Project
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E-Mail with your Background
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Anyone feeling underwhelmed or overwhelmed
with course?
Name & General Description of Project
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e.g. "Hannay Reels"
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Manufactures Steel & Aluminum Reels for Hose & Cable
in Industrial Applications
Applications: Customer Relations Management, Order
Entry, Accounts Receivable, Purchasing, Accounts
Payable, Inventory, Manufacturing Control, Payroll, etc.
Sample Projects: http://cs.union.edu/csc340
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Example of a Relation
attributes
(or columns)
tuples
(or rows)
Attribute Types
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The set of allowed values for each attribute
is called the domain of the attribute
Attribute values are (normally) required to
be atomic; that is, indivisible
The special value null is a member of
every domain
The null value causes complications in the
definition of many operations
Relation Schema and Instance
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A1, A2, …, An are attributes

R = (A1, A2, …, An ) is a relation schema
Example:
instructor = (ID, name, dept_name, salary)
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Formally, given sets D1, D2, …. Dn a relation r is a subset of
D1 x D2 x … x Dn
Thus, a relation is a set of n-tuples (a1, a2, …, an) where each ai
 Di
 The current values (relation instance) of a relation are specified
by a table
 An element t of r is a tuple, represented by a row in a table
Relations are Unordered
 Example: instructor relation with unordered tuples
 Order of tuples is irrelevant (tuples may be stored in an arbitrary order)
Database
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A database consists of multiple relations
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Information about an enterprise is broken up into parts
instructor
student
advisor
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Bad design:
univ (instructor -ID, name, dept_name, salary, student_Id, ..)
results in
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repetition of information (e.g., two students have the same instructor)
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the need for null values (e.g., represent an student with no advisor)
Normalization theory (Chapter 7) deals with how to design “good”
relational schemas
Keys
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Let K  R
K is a superkey of R if values for K are sufficient to identify a
unique tuple of each possible relation r(R)
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Superkey K is a candidate key if K is minimal
Example: {ID} is a candidate key for Instructor
One of the candidate keys is selected to be the primary key.
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Example: {ID} and {ID,name} are both superkeys of instructor.
which one?
Foreign key constraint: Value in one relation must appear in
another
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Referencing relation
Referenced relation
Schema Diagram for University Database
Relational Query Languages
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Procedural vs.non-procedural, or declarative
“Pure” languages:
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Relational algebra
Tuple relational calculus
Domain relational calculus
Relational operators
Selection of tuples
 Relation r
 Select tuples with A=B
and D > 5

σ
A=B and D > 5
(r)
Selection of Columns (Attributes)

Relation
r
 Select A and C

Projection

Π
A, C
(r)
Joining two relations – Cartesian Product
 Relations r, s
 r xs
Union of two relations

Relations r, s
 r s
“Reel” Life Example of “Union” [1 of 2]
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“Reel” Life Example of “Union” [2 of 2]
The Query
(UPR00100=PayrollNameAddr; UPR00400=PayMaster; UPR00500=DeductionMaster)
SELECT
FROM
WHERE
UPR00100.EMPLOYID, UPR00100.EMPLCLAS, UPR00100.LASTNAME, UPR00100.FRSTNAME, UPR00100.MIDLNAME,
UPR00100.BRTHDATE, UPR00100.GENDER, UPR00100.STRTDATE, UPR00400.PAYRCORD, UPR00400.PAYRTAMT,
UPR00500.DEDUCTON
UPR00100 INNER JOIN
UPR00400 ON UPR00100.EMPLOYID = UPR00400.EMPLOYID LEFT OUTER JOIN
UPR00500 ON UPR00100.EMPLOYID = UPR00500.EMPLOYID
(UPR00400.PAYRCORD = 'WEEKLY' OR
UPR00400.PAYRCORD = 'HOURLY') AND (UPR00100.EMPLCLAS IN (@emplClass)) AND
(UPR00500.DEDUCTON IN (@wchMED))
UNION
SELECT
FROM
WHERE
UPR00100_1.EMPLOYID, UPR00100_1.EMPLCLAS, UPR00100_1.LASTNAME, UPR00100_1.FRSTNAME, UPR00100_1.MIDLNAME,
UPR00100_1.BRTHDATE, UPR00100_1.GENDER, UPR00100_1.STRTDATE, UPR00400_1.PAYRCORD, UPR00400_1.PAYRTAMT,
'NONE' AS DEDUCTON
UPR00100 AS UPR00100_1 INNER JOIN
UPR00400 AS UPR00400_1 ON UPR00100_1.EMPLOYID = UPR00400_1.EMPLOYID
(UPR00400_1.PAYRCORD = 'WEEKLY' OR
UPR00400_1.PAYRCORD = 'HOURLY') AND (UPR00100_1.EMPLCLAS IN (@emplClass))
AND (NOT EXISTS
(SELECT
EMPLOYID
FROM
UPR00500 AS UPR00500_1
WHERE
(EMPLOYID = UPR00100_1.EMPLOYID) AND (DEDUCTON LIKE 'MED%')))
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Set difference of two relations

Relations r, s
 r –s
Set Intersection of two relations
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Relation r, s
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rs
Joining two relations – Natural Join
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Let r and s be relations on schemas R and S
respectively.
Then, the “natural join” of relations R and S
is a relation on schema R  S obtained as
follows:
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Consider each pair of tuples tr from r and ts from s.
If tr and ts have the same value on each of the
attributes in R  S, add a tuple t to the result,
where
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t has the same value as
t has the same value as
tr on r
ts on s
Natural Join Example
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Relations r, s
 Natural Join

r
s
Figure in-2.1
MySQL Workbench
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Handout
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Relationships between tables in
HannayReels sample Customer Tables
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"Live" "Reel"
Database Example
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email automatically sent today
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Draw relation between Invoices table and
Payments table on board
Show Query in Reporting Services
FIX problem on Tuesday using SQL
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Homework/Project
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Homework due Next Class:
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Homework due in One Week:
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1.1, 1.3, 1.5, 1.6
2.1, 2.2, 2.4, 2.6, 2.7, 2.8
Project due Next Class:
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Name of Your Enterprise
General Description
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To the Computers
“All your base are belong to us”
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http://en.wikipedia.org/wiki/All_your_base_are_belong_to_us
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Examining Relations (Customers)
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In OpenOffice “BASE”
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In MS Access
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Tools Menu  Relationships
“Database Tools” Tab  “Relationships” button
Start Creating a few Tables for Your DB
In MySQL Workbench (see HannayReels.mwb)
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If you are unsure on project, try a few
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