Lecture 2dfavfbagfbv

7/28/2019 Lecture 2dfavfbagfbv

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Database ManagementSystems II

Lecture 2


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Last Lecture Introduction to unit

Database Design Process

Conceptual Modeling Enhanced ER Model


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Any questions??? Class exercise?

Practical 1?

Tutorial 1?


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This Lecture Logical Database Design

Relational Model

E-ER Relational Mapping


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Relational Model - Review The basic construct of the relational

model is the relation.

A relation consists of

relation schema

relation instance


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Relational Model Review

(contd.) Relation Schema = relation name

+ {name of each field: domain of

each field}

COURSE (course_no: integer, title:

string, credits:integer, year:integer,semester:integer)

Relation Instance = set of tuples


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(contd.) Example


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(contd.)Arelational database is a collection

of relations with distinctrelation names

Integrity constraint is a rule (orcheck) that is enforced by the DBMS toensure that incorrect data is not storedin the database


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(contd.) Types of Integrity Constraints:

Domain constraints (domain of a field)

Key Constraints (UNIQUE & PRIMARYKEY)

Foreign Key Constraints (FOREIGN KEY)

Table Constraints (constraints withintables)

Assertions (constraints between multipletables)


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(contd.) Domain constraints: specify that each

attribute A must be an atomicvalue from the

domain of A. Atomic value: That is, the value is non

divisible into components

Therefore, relational model doesnt allow multi-

valued attributes or composite attributes!

Relational database provides data types tospecify valid domains.


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(contd.)Key Constraints (UNIQUE & PRIMARY

KEY)

The minimal set of attributes thatuniquely identify a tuple is called thekey of a relation

There may be multiple candidate keysin a relation, of which one is chosen asthe primary key.

Primary key is underlined


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(contd.)

LicenseNumber EngineSerialNumber

Make Model Year

32-3789 12342 Toyota Corolla 2000

17-2489 22495 Nissan Sunny 1989

54-5680 54356 Toyota Vannette 1990

WP-1268 24322 Mitsubishi Galant 2001

*

VEHICLE

Candidate keys, Super keys, Primary Key


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(contd.) Foreign Key Constraints

We can designate an attribute(s) as a

foreign key(s)

Foreign key(s) attribute(s) always refer to

attribute(s) of the same or another entity

Foreign keys enforce referential integrity

constraints


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(contd.)Foreign key attributes in R1 referring to R2 have the

following rules:

The FK attributes in R1 have the same domain(s)asthe primary key attributes if R2 The value of FK in tuple t1 in R1 must reference an

existingPK value in tuple t2 of R2

We can diagrammatically display the foreign keys by

drawing an arrow from the foreign key to the primarykey


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(contd.)

Referential Integrity Constraints displayed in the relational schema


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Relational Model Review(contd.)

Many relational DBMSs support most of the constraintsdescribed above.

However, many applications require more advanced features for

specifying integrity constraints.

Table Constraints (constraints within tables) Balance of the account should be greater than 0

In SQL, CHECK constraint can be used

Assertions (constraints between multiple tables) For example: An employees salary should not exceed his/her

managers salary

Such constraints can be using specific language features such astriggers and assertions


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Logical Database Design -Revisited

Logical Database Design is the process

of converting the E-ER Model to theRelational Model

There are a set of conversion rules thatcan be applied


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E-ER to Relational Mapping(Review)

E-ER Model RelationalModel

Entity Relation

For example,STUDENT STUDENT


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E-ER to Relational Mapping(Review contd.)


Simple Attributes Attributes

For example,

STUDENT

name

nameSTUDENT


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Primary Key Primary Key

For example,

STUDENT

name

nameSTUDENT


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E-ER Model Relational Model

Composite attributes Set of simple

attributes

For example,

EMPLOYEE

fullname

surname

firstname

first-name

sur-name

EMPLOYEE


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1:1 relationship Foreign key

relationship

For example,

EMPLOYEE

ADDRESShas

id

a

idEMPLOYEE

aeidADDRESS


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Usually, if there is a 1:1 relationship Rfrom entity A to B and if B is in total

participation with A on R then theforeign key is placed in B

For example,

A

B

R

A

B


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If A and B are both in total participation with

R then A & B are collapsed as 1 table

For example,

A

BR

AB


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1:N Foreign key

relationship

For example,

ARTIST

OBJECTScreates

id

yr_

created

id

yr_createdid

ARTIST

OBJECT


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M:N Relationship

relation and twoforeign keys

ARTIST

ART-OBJECTcreates

name

Date-

Created

id

name

id

id Datecreated

nameCREATES

ART_OBJECT

ARTIST


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Multivalued attribute Relation &Foreign Key

For example,

STUDENT

middle namesid

id

middle-nameid

STUDENT

MIDDLE_NAMES


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N-ary relationship Relationshiprelation and nforeign keys

For example,

A

C

B

D

r

pkC

pkA pkB

pkD

pkDpkCpkBpkA

A B

R

C

D

E-ER to Relational Mapping

(Review contd.)


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E-ER to Relational Mapping(contd.)


Subclass/superclass Multi relation options:

- Option 1- Option 2

Single relation options:

- Option 3- Option 4


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C is a super class with primary key, PK(C) = k

Attributes of C = Attrs(C) = {k, a1, , an}

S1

, S2

, , Sm

are subclasses of C

C

k a1 a2 an

S1 S2 Sm

ISA


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Option 1:

Create a relation L for superclass C suchthat Attrs(L) = {k, a1, a2, , an} andPK(L) = k

For each Si, create a relation Li suchthat Attrs(Li) = {k} U Attrs(Si) andPK(Li) = k


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Relation L for class C

Relation Li for each subclass Si

Foreign key link from PK(Li) to PK(L)

EQUIJOIN between Li and L to obtain all attributes ofLi

Option 1 works for all constraints disjoint,overlapping, total and partial.


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Option 2:

For each subclass Si, create a newrelation Li such that Attrs(Li) = Attrs(C)U Attrs(Li) and PK(Li) = k


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The ISA relationship must be total (i.e.subclasses must cover the super class)

If the subclasses are not disjoint, wekeep multiple copies of Cs attributesvalues

To obtain all entities in superclass C, weneed to perform a full outer join withall relations Li


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Option 3:

Create a single relation L with Attrs(L) =Attrs(C) U Attrs(S1) U U Attrs(Sm) U {t}such that PK(L) = k where t is type specifyingwhich subclass the entity belongs if any.

The specialization/generalization relationshipmust be disjoint


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Good if subclasses have fewattributes

No EQUIJOIN or OUTER JOIN operationneeded

If subclasses have a lot of attributes,potential for many null values


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Option 4:

Create a single relation L with Attrs(L) =Attrs(C) U Attrs(S1) U U Attrs(Sm) U {t1, t2,, tm} such that PK(L) = k where ti is aBoolean attribute indicating whether each

tuple belongs to subclass Si.

This relation allows overlapping constraintsfor specialization/generalization relationship


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When having multilevel specialization(or generalization) hierarchy, we do not

need to follow the same mappingoption.

We can use different mapping optionsfor different parts of the hierarchy


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Aggregation

Relationship has M:N cardinality

Entity A Entity BR1

R2 Entity C

PKA

A

PKB

B

PKC

C

PKBPKA

R1

PKB PKcPKA

R2


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Relationship has 1:M cardinality

Entity A Entity BR1

R2 Entity C

PKA

A

PKAPKB

B

PKC

C

PKB PKc

R2


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Summary

Relational Model

Logical Database Design

Enhanced ER Model Relational Model

Practicals and Tutes will cover exercisesin these topics

Documents

Lecture 2dfavfbagfbv