ER/EER to Relational Data Model 1 Database Design

ER/EER to Relational Data Model

1

Database Design

Database DesignDatabase Design

2

Steps in building a database for an application:

Real-world domain

Conceptualmodel

DBMS data model

Create Schema

(DDL)

Load data(DML)

Relational ModelRelational Model

3

It uses the concept of mathematical relation

In the relational model, all data is logically structured within relations (tables).

Relational Data StructureRelational Data Structure

4

4170010

4182000

4182034

4188134

4189860

4192134

StudentNo

Al-Saleh

Al-Ghanem

Al-Fahad

Saod

Rashed

Al-Fahad

LName

Amal

Nora

Laila

Amal

Rana

Rania

FName Initial

M.

A.

A.

F.

I.

M.

DOB

04-06-78

02-12-79

01-11-74

22-04-73

30-01-78

19-03-79

GPA

3.91

4.20

4.01

3.01

2.31

3.50

STUDENT

Attributes

Tuples

Cad

inal

ity

Rel

atio

n

Degree

Relation’s name


5

Relation is a table with columns & rows. Holds information about entities.

Attribute is a named column of a relation.

Domain is the set of allowable values for one or more attributes. Every attribute in a relation is defined on a domain.

Tuple is a row of a relation.


6

Degree of a relation is the number of attributes it contains.

Cardinality of a relation is the number of tuples it contains.

Relational database is a collection of normalized relations with distinct relation names.

Tuples

7

A tuple is a set of (<attribute,value>) pairs.These tuples are identical:

T1 = <(name, ahmad), (age, 16), ( address, Riyadh)>

T2= <(age, 16), ( address, Riyadh),(name, ahmad)>

The ordering of the tuples is not part of the relation.

Domain

8

Domain : set of atomic valuesDomain Constraints : values of each

attributes must be an atomic (not divisable into components ) value from the domain.

Associated to a data type.Dom(colors): {black,red, white},

DomainsDomains

9

4170010

4182000

4182034

StudentNo

Al-Saleh

Al-Ghanem

Al-Fahad

LName

Amal

Nora

Laila

FName Initial

M.

A.

A.

DOB

04-06-78

02-12-79

01-11-74

GPA

3.91

4.20

4.01

STUDENT

StudentNo

LName

FName

Initial

Student Name

Last Name

First Name

Initial

DefinitionAttribute

DOB

GPA

Domain Name

Date of Birth

Great Point Average

Digits: size 7

Character: size 15

Character: size 15

Character: size 3

Date: range 01-01-20, format dd-mm-yy

Real: size 3, decimal 2, range 0-5

Relational KeysRelational Keys

10

Candidate key (CK) It’s a minimal set of attributes necessary to identify a row. Its also called minimal super key.A relation schema may have more than one key, each of the keys is called a candidate key.

Primary Key (PK) is the CK that is selected to identify tuples uniquely within the relation

Foreign Key (FK) is an attribute, or set of attributes, within one relation that matches the CK of some relation. Used to represent relationship between tuples of two relations.

Relational KeysRelational Keys

11

4170010

4182000

4182034

4188134

4189860

StudentNo

Al-Saleh

Al-Ghanem

Al-Fahad

Saod

Rashed

LName

Amal

Nora

Laila

Amal

Rana

FName Initial

M.

A.

A.

F.

I.

DOB

04-06-78

02-12-79

01-11-74

22-04-73

30-01-78

GPA

3.91

4.20

4.01

3.01

2.31

STUDENT

D001

D002

D003

DeptNo

Computer Science

Business Administration

Science

Department Name

Build # 20

Dept

D001

D001

D002

D003

D001

Location

Build # 45

Build # 6

DEPARTMENTPrimary Key

Foreign Key

Basic Definitions

12

Relational Keys:PK : underlined with a straight line Composite keysFK: underlined with a dashed line

Properties of tables:Each table has a unique nameEach entry stores a single valueEach row is uniqueEach column has a unique nameThere is no sequence of rows.

DB RelationsDB Relations

13

Relation schema is a named relation defined by a set of attributes

If A1, A2, .., An are a set of attributes, then relation schema R is:

R = (A1, A2, .., An)Relational database schema is a set of

relation schemas, each with a distinct name If R1, R2, .., Rn are a set of relation schemas,

then relational schema R is:R= {R1, R2, .., Rn}

Relation SchemaRelation Schema

14

STUDENT (StudentNo, Lname, Fname, Initial, DOB, GPA, Dept)

4170010

4182000

4182034

4188134

4189860

StudentNo

Al-Saleh

Al-Ghanem

Al-Fahad

Saod

Rashed

LName

Amal

Nora

Laila

Amal

Rana

FName Initial

M.

A.

A.

F.

I.

DOB

04-06-78

02-12-79

01-11-74

22-04-73

30-01-78

GPA

3.91

4.20

4.01

3.01

2.31

STUDENTDept

D001

D001

D002

D003

D001

Relation SchemaRelation Schema

15

DEPARTMENT (DeptNo, Department Name, Location)

D001

D002

D003

DeptNo

Computer Science

Business Administration

Science

Department Name

Build # 20

Location

Build # 45

Build # 6

DEPARTMENT

Relational IntegrityRelational Integrity

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Data integrity refers to the validity, consistency, and accuracy of the data in the database.

Integrity rules are constraints that apply to all instances of the DB.

Two integrity rules for the relational model:- Entity integrity- Referential integrity

Entity Integrity

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Entity Integrity: (in a single relation)Ensures that there are no duplicate records within the table. In a base relation, no attribute of a PK can be null.T[PK] ≠ Null for any tuple in R.

Referential Integrity

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Referential Integrity: ( two relations) If a FK exists in a relation, either the FK value

must match a CK value of some tuple in its home relation or the FK value must be wholly null.

The referencing relation R1 has the FK that reference a PK in the referenced relation R2.

=> t1[FK] =t2[PK], t1 in R1 and t2 in R2

Referential Integrity Constraints

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SSN | Fname | LName | Bdate | DNO

Dnumber | Dname | MGRSSN | MgrStartDate

EMPLOYEE

DEPARTMENT

Properties of Relations

20

1. The relation has a distinct name from all other relation names in the relational DB (oracle enforces this rule)

2. Each cell of the relation (attribute) contains exactly single value

3. Each attribute has a distinct name (oracle enforces this rule)

4. The values of an attribute are all of the same domain5. Each tuple is distinct. There are no duplicate tuples6. The order of attributes has no significance7. the order of tuples has no significance; theoretically.

Integrity constraints

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Domain constraintsEntity integrity constraintsReferential integrity constraintsKey ConstraintOperational constraints

Operations of the Relational Model

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Updates: changes the database’s state.InsertDeleteUpdate/modify

Retrievals: does not change the current state of the database.

The Insert Operation

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This operation can violate all constraints of the relation. How?

Domain constraint: attribute value does not apear in the corresponding domain.

Key constraint: if the key value already exist in another tuple.

Entity integrity: if the PK of the new tuple is NULLReferential integrity: if any value of the FK refers to

a tuple that does not exist in the other relation.

The Delete Operation

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This operation can violate only the referential integrity constraint. How?

E.g: you can not delete a course that many students study in already.

The Update Operation

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The new value must be of the correct data type and domain.

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Step -1- Strong Entity Type

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Represent each entity type with a relationEntity type attributes become the relation

attributese.g.

STUDENT (StudentNo, Lname, Fname, Initial, DOB, GPA, Dept)DEPARTMENT (DeptNo, Department Name, Location)

Step -2- Weak Entity Type

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A weak entity type relation must include its key and its strong entity type PK as a FK. The combination of the two keys form the PK of the weak entity.

EMPLOYEE

EmpNo

LName FName

DOBDepNo FName

DEPENDENThas

EMPLOYEE (EmpNo, Lname, Fname, DOB)

DEPENDENT (DepNo, EmpNo, FName)FK

Step -3- Binary 1:1 Relationship Type

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Identify an entity type (S) (preferably total participator) Include the PK of the other entity (T) as a FK in S Add attributes that describes the relationship

BRANCHmanageEMPLOYEE

EMPLOYEE(EmpNo, Lname, Fname, DOB)

BRANCH(BrnNo, Name, EmpNo, StartDate, EndDate)

1 1

(0,1) (1,1)

StartD EndDEmpNo

LName FName

DOBName

BrnNo

FK

………..

Binary 1:1 Important Notes

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For a 1:1 relationship, the PK of the partial is a FK in the total.

What happens if have an equal multiplicity?Then it depends on the path analysis.

Step -4- Binary 1:N Relationship Type

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Identify a participating entity type (S) on the m-side Include the PK of the other entity type (T) as a FK in S Add attributes that describes the relationship

BRANCHallocateEMPLOYEE

EMPLOYEE(EmpNo, Lname, Fname, DOB, BrnNo)

BRANCH(BrnNo, Name)

M 1

EmpNo

LName FName

DOBName

BrnNo

FK………..

Binary 1:N Important Notes

32

For a 1:N relationship, the schema of the N side will have the PK of the 1 side as a foreign key.

What if the 1 side has a composite key?All keys will be copied to the many side as

foreign keys.

Step -5- Binary M:N Relationship Type

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Create a relation R to represent the relationship Include the PK of participating entity types (T & S) as FK in R. The combination of the two FK will form the PK of R Add attributes that describes the relationship

EMPLOYEE(EmpNo, Lname, Fname, DOB)

PROJECT(ProjNo, Name)

Work-on(EmpNo,ProjNo, hours)

M N

hoursEmpNo

LName FName

DOBName

ProjNo

FKFK

EMPLOYEE work-on PROJECT

Binary M:N Important Notes

34

For a M:N relationship , a new schema is created that contains the PK of N and PK of M plus any other attributes.

Some times the situation of insufficient PK occurs … !! How?

E.g. a patient gets a treatment several times a resulting table will contain the PK of patient and the PK of the treatment.. But the patient may have the same treatment for several times ….. Therefore a date field must be included as a PK of the relation.

Step -6- n-ary Relationship Type

35

Create a relation R to represent the relationship Include the PK of the participating entities as FK in R. The combination of all FK form the PK of R. Add attributes that describes the relationship

SUPPLIERcontractBUSINESS

BUSINESS(BizNo) LAWYER(LawNo) SUPPLIER(SupNo)

contract(BizNo, SupNo, LawNo, StartDate, EndDate)

StartD EndDBizNo

LawNo

SupNo

FK

Lawyer

FK FK

Ternary Relationship Important Notes

36

If the Ternary relationship is:N:N:N then a new intersection table is

created that includes the 3 entities keys as a PK of that table.

1:N:N then the PK of the 1 is a FK in the intersection table.

Recursive Relationship 1:1

37

The schema for this relationship is going to have a new field called MarriedToPersonID, which shares the same domain of the PersonID.

PERSON Married to

PersonID1

1

PERSON(PersonID, PersonName,MarriedToPersonID)

FK

Recursive Relationship 1:N

38

The schema for this relationship is going to have a new field, Its exactly the same is a 1:1 uniary relationship

EMPLOYEE supervises

EmpID1

N

EMPLOYEE(EmpD, EmpName,DOB, ManagerEmpID)

FK

Step-7- Multi-valued Attribute

39

Suppose A is a relation that contains the multi-valued attribute Create a relation R to represent the attribute Include the PK of A as FK in R The PK of R is the combination of the PK of A (FK) & the multi-valued attribute

EMPLOYEE

EmpNo

DOB

Tel_no

EMPLOYEE(EmpNo, DOB)

TELEPHONE(EmpNo, tel_no)FK

Composite Attribute

40

Include its simple components in the relation

EMPLOYEE

emp_no

name DOB

LNameinitial FName

EMPLOYEE(EmpNo, Fname, initial, Lname, DOB)

41

EER To Relational Data Model

EER Relational Model

42

In EER Model, there are sub-classes and super-classes and to convert them into relational schema we have to use one of the four options which will discuss in next slides:

We have to remember these concepts which related to EER.

o Total : all entities in super class must be a member of subclass.

o Partial :an entity in super class may not belong to any subclass.

o Disjoint : an entity be a member only in one subclass.

o Over-lapping: an entity may be a member of more than one subclass

EER Relational ModelMandatory(Total)/NonDisjoint(Overlapping)

43

Suppose specialization with subclasses (S1, S2, .., Sm} & a

superclass C Create a relation L to represent C with PK

& attributes Include the unshared attributes for each

subclass Si, 1<= i <= m Add discriminator to distinguish the type of

each tuple

EER Relational ModelTotal/Non-Disjoint (Overlapping)

Single Relation

R

PK(K)

Boolean type attribute

o

EER Relational ModelTotal/Non-Disjoint (Overlapping)

45

EMPLOYEE( EmpNo, Fname, Lname, DOB, Salary,TypingSpeed,TGrade,EngType, Secretary Flag, Technician Flag, Engineer Flag)

EER Relational ModelTotal/Disjoint

46


superclass C Create a relation Li, 1≤ i ≤m, to represent

each combination of super/subclass


Relation R2 for subclass

TRUCK

Relation R1 for subclass

CAR

PK(K)

Tonnage

d


d

Car( VehicleId ,LicensePlateNo, Price, MaxSpeed ,NoOfPassengers)

Truck ( VehicleId ,LicensePlateNo, Price,NoOfAxles,Tonnago)

EEROptional/NonDisjoint(Overlapping)

49


superclass C Create a relation L to represent C with PK

& attributes Create a relation Li to represent each

subclass Si, 1≤ i ≤m

Relation R1


Relation R

Relation R2

Relation R3

PK(K)O


O

Employee( SSN ,Fname,Minit,Lname,BirthDate,Address,JobType)Secretary( SSN ,TypingSpeed)Technician( SSN , Tgrade)Engineer( SSN ,EngType)

EEROptional/Disjoint

52


superclass C Create a relation L1 to represent C with PK

& attributes Create a relation L2 to represent all

subclasses Si, 1≤ i ≤ m Add discriminator to distinguish the type of

each tuple

Relation R1


Relation R

Relation R2

Relation R3

PK(K)d


54

EMPLOYEE ( EmpNo, Fname, Lname, DOB, Salary)SECRETARY( EmpNo, TypingSpeed)TECHNICIAN( EmpNo, Tgrade)ENGINEER( EmoNo, EngType)

Views

55

Base relation is a named relation corresponding to an

entity in the conceptual schema, whose tuples are physically stored in the DB

View is a derived relation. Virtual, may not exist, but dynamically derived from one or more base relations.

The only information about a view that is stored in the database is its structure.

The external model can consist of both conceptual level relations (base relations) and derived views.

View

56

Purpose of Views

57

Provides security mechanism by hiding parts of the DB from certain users

Customize data to user’s needs, so that the same data can be seen by different users in different ways

Simplify complex operations. It allow you to work with data from different tables simultaneously.

Supports logical data independence

Updating Views

58

All updates to a base relation should be immediately reflected in all views that reference that base relation.

All updates to a view should be reflected in the underlying base relation, under the following rules:- Updates are allowed through a view defined using a

simple query involving a single base relation & containing either the PK or a CK of the base relation

- Update are NOT allowed through views involving multiple base relations

- Updates are not allowed through views involving aggregation or grouping operations

Relational Integrity

59

Data integrity refers to the validity, consistency, and accuracy of the data in the database.

Integrity rules are constraints that apply to all instances of the DB.

Two integrity rules for the relational model:- Entity integrity- Referential integrity

Relational Integrity

60

Entity Integrity: Ensures that there are no duplicate records within the table. In a base relation, no attribute of a PK can be null

Referential Integrity: If a FK exists in a relation, either the FK value must match a CK value of some tuple in its home relation or the FK value must be wholly null.

Enterprise constraints: rules specified by the users or DBA of the DB based on the ways an organization perceives and uses its data

(e.g.number of staff working in a branch is at most 20)

References

61

Elmasri,Ramez &Navathe,Shamkant B., Fundamentals Of Database System, fifth edition, Pearson ,2007

Fatimah Alakeel’s Lectures http://webdocs.cs.ualberta.ca/~drafiei/291/notes/4.

ER2Rel.pdfhttp://www.cs.sfu.ca/CC/354/zaiane/material/notes/

Chapter2/node10.html#SECTION00190000000000000000

http://web.uettaxila.edu.pk/CMS/seADMSbsSp09/notes%5CADBMS-Lecture-6%20ER%20to%20Relational%20Mapping.pdf

www.cs.gsu.edu/~cscbecx/.../Chapter%207.ppthttp://www.mscs.mu.edu/~praveen/Teaching/

Fa06/Db/Lectures/ch7.pdf

www.cs.uwm.edu/classes/cs557/powerpoint/chapter9.ppt

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ER/EER to Relational Data Model 1 Database Design