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I think Answer should be 3, but given is 4..

No of table will be 3

E1, E2,E3 we need 3 tables

but as it gives 1:n relationship , so no more tables are needed
by Veteran (118k points)
selected
0
Please mention the primary key and foreign key(if any) of each of the tables.
0
E3 will be saparate  with E is key
First E1 R will combine with key CG then E1 R P will combine with key E1 G
E2 will kept seperate with key G.
0
Primary key of table E3 i.e. E works as a foreign key in the relationship set E1PR. And primary key of E1PR is CG.

Is that correct?
While converting ER Diagram to Relation Table, We create a separate table for:

1. WEAK Entity.
2. STRONG Entity.
3. M:N Relationship.
4. Composite Attribute.
5. Multivalued Attribute.

by Junior (911 points)
0
We don't create seperate table for composite attributes.
I think the answer must be 4.

For E1,E2,E3 we will defiantly need one table each.

By putting primary key of E2 i,e G in E1's table we can specify the indentifing relationship R between them. Hence no need of table for R.

But For P we will need a different table in which we will put primary key of E3 i,e E and Primary key of E1 i,e C  as foreign keys and the primary key of the table for  P will be the combination of E and  C i,e EC

Hence we need total 4 tables( for E1, E2 ,E3 and for P)
by Active (3.3k points)

+1 vote