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35 votes
35 votes

The following table has two attributes $A$ and $C$ where $A$ is the primary key and $C$ is the foreign key referencing $A$ with on-delete cascade.
$$\begin{array}{|c|c|} \hline \textbf {A} &  \textbf {C} \\\hline \text {2} &  \text{4} \\\hline \text{3} & \text{4} \\\hline \text{4} & \text{3} \\\hline \text{5} &  \text{2} \\\hline \text {7} &  \text{2} \\\hline \text{9} & \text{5} \\\hline \text{6} & \text{4} \\\hline \end{array}$$

The set of all tuples that must be additionally deleted to preserve referential integrity when the tuple $(2, 4)$ is deleted is:

  1. $(3, 4)$ and $(6, 4)$
  2. $(5, 2)$ and $(7, 2)$
  3. $(5, 2), (7, 2)$ and $(9, 5)$
  4. $(3, 4), (4, 3)$ and $(6, 4)$
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4 Comments

simple and grt explanation @Bikram sir !
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Thanks Sir for the explanation.
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Thank You, sir  you explained very well
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2 Answers

46 votes
46 votes
Best answer

(C)
Since deleting $(2,4)$, since $2$ is a primary key, you have to delete its foreign key occurence i.e $(5,2)$ and $(7,2)$
Since we are delting $5$, and $7$ we have delete it foreign key occurence i.e $(9,5)$.

There is no foreign key occurence for $9$.

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1 comment

Very good Question and explanation is extremely well

I understand On delete cascade in single Table
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13 votes
13 votes

here A is behaving like parent and C is  behaving like child bcoz A is PK and C is FK referencing to A.

when (2,4) is deleted then it force to delete all entries in which C contains 2 bcoz there is no 2 remains in parent so child not able to access 2. so delete (5,2) (7,2) which force to delete all entries in which C contains 5or7 bcoz there is no 5 or 7 remain in parent so child not able to access 5 or 7.so delete (9,5) which force to delete all entries in which C contains 9 .

so ans should be C

1 comment

Nice explanation.Easy to understand on delete cascade  for single table.
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