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CSE 880: Database Systems

Lecture : EER to Relational Mapping

Reference:

Read Chapter 9 of the textbook

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EER to Relation Mapping

Mapping of superclass/subclass relationships Mapping of shared subclasses Mapping of union types (categories)

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Mapping of Superclass/Subclasses

Let R be the superclass {S1, S2,….,Sm} are the subclasses

– Each subclass can have its own local attributes

R

S1 S2 Sm

…

ka1

a2

b1

b2

c

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EER to Relation Mapping

Step 8: 4 possible approaches– Option 8A: Multiple relations – Superclass and subclasses

– Option 8B: Multiple relations – Subclass relations only

– Option 8C: Single relation with one type attribute

– Option 8D: Single relation with multiple type attributes

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Mapping EER Constructs to Relations

Option 8A: Multiple relations - superclass and subclass– Create superclass relation with attributes {k,a1,a2} and primary key

(PK) = k

– Create a relation Si for each subclass Si, with attributes {k} U {attributes of Si} with PK = k.

R

S1 S2 Sm

…

ka1

a2

b1

b2

c1

k a1 a2

R

k c1 c2

Smc2

k b1 b2

S1

…

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Example

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MySQL Workbench Example

Click on 1-1 identifying relationship type

Click on subclass first before superclass. Repeat this for all 3 subclasses.

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MySQL Workbench Example

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Mapping EER Constructs to Relations

Option 8B: Multiple relations –subclass relations only– Create a relation for each subclass Si, with the attributes of Si

and the superclass

– Works for total specialization (i.e., every entity in the superclass must belong to (at least) one of the subclasses)

R

S1 S2 Sm

…

ka1

a2

b1

b2

c1

c2

a1 a2k c1 c2

Sm

a1 a2k b1 b2

S1

…

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Example

Tonnage

Total specialization

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Limitation of previous two approaches

Query: What does John Doe do as an employee?

To answer this query, we need to scan all three subclass relations, which is inefficient!

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Option 8C: Single relation – with one type attribute (t)– Create a single relation with attributes {k,a1,…an} U {attributes of

S1} U…U {attributes of Sm} U {t} with primary key, PK = k

– The attribute t is called a type (or discriminating) attribute

– This option works for disjoint specialization

EER to Relations Mapping

d

C

S1 S2 Sm

…

ka1

a2

b1

b2

c1

c2

t a1 a2k b1 b2

R

c1 c2… t

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Example

EngType

d

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EER to Relations Mapping

Option 8D: Single relation – with multiple type attributes– Create a single relation with attributes {k,a1,…an} U {attributes of

S1} U…U {attributes of Sm} U {t1, t2,…,tm} and PK = k

– ti is a Boolean attribute indicating whether a tuple belongs to Si.

– This option works for overlapping specialization

o

C

S1 S2 Sm

…

ka1

a2

b1

b2

c1

c2

a1 a2k b1 b2

R

c1 c2… …t1 tm

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Example

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So which option is better?

Depends on applications; must consider tradeoff:– Too many relations (options 8A and 8B)

Inefficient query processing

– Single relation (options 8C and 8D) Lots of nulls; may “lose” some meaningful relationships

If everything is mapped to the Employee relation, we lose

the relationship between hourly employee and trade

union

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EER to Relations Mapping

Mapping of Shared Subclasses (Multiple Inheritance)– A shared subclass is a subclass of several superclasses,

indicating multiple inheritance. These superclasses must have the same key attribute

– Can apply any of the four options (subject to their restrictions – total/partial, overlapping/disjoint)

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o

Example

STUDENT_ASSISTANT is a shared subclass of the EMPLOYEE and STUDENT entity types

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Example

Since there are usually separate queries for employees, alumni, and students, we can use options 8A or 8B

– Relations for option 8A: Person, Employee, Alumnus, Student

– Relations for option 8B: Employee, Alumnus, Student

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Example

o

Suppose we want to use option 8C and 8D to group all employees into a single relation called EMPLOYEE

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Example

o

Suppose we want to use option 8C to group all students into a single relation called STUDENT

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Putting it all together…