SQL
October 4, 2023SQL
October 4, 2023SQL
| Question 36 |
Consider a database that has the relation schema EMP (EmpId, EmpName, and DeptName). An instance of the schema EMP and a SQL query on it are given below.
The output of executing the SQL query is ___________.
| 2.6 | |
| 2.7 | |
| 2.8 | |
| 2.9 |
Question 36 Explanation:
The given query is
⇾ We start evaluating from the inner query.
The inner query forms DeptName wise groups and counts the DeptName wise EmpIds.
⇾ In inner query DeptName, Count(EmpId) is the alias name DeptName, Num.
So, the output of the inner query is,
The outer query will find the
Avg(Num) = (4+3+3+2+1)/5 = 2.6
⇾ We start evaluating from the inner query.
The inner query forms DeptName wise groups and counts the DeptName wise EmpIds.
⇾ In inner query DeptName, Count(EmpId) is the alias name DeptName, Num.
So, the output of the inner query is,
The outer query will find the
Avg(Num) = (4+3+3+2+1)/5 = 2.6
Correct Answer: A
Question 36 Explanation:
The given query is
⇾ We start evaluating from the inner query.
The inner query forms DeptName wise groups and counts the DeptName wise EmpIds.
⇾ In inner query DeptName, Count(EmpId) is the alias name DeptName, Num.
So, the output of the inner query is,
The outer query will find the
Avg(Num) = (4+3+3+2+1)/5 = 2.6
⇾ We start evaluating from the inner query.
The inner query forms DeptName wise groups and counts the DeptName wise EmpIds.
⇾ In inner query DeptName, Count(EmpId) is the alias name DeptName, Num.
So, the output of the inner query is,
The outer query will find the
Avg(Num) = (4+3+3+2+1)/5 = 2.6
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