Question at position 13 The director of the Wellesley Recreational Center would like to know the number of student registrations for each class offered. What should be in A and B below in the query design? A is Count and B is Group ByA is Group By and B is CountA is Group By and B is SumA is Sum and B is Group By单项选择题
A
A is Count and B is Group By
B
A is Group By and B is Count
C
A is Group By and B is Sum
D
A is Sum and B is Group By

Question

Question at position 13 The director of the Wellesley Recreational Center would like to know the number of student registrations for each class offered. What should be in A and B below in the query design? A is Count and B is Group ByA is Group By and B is CountA is Group By and B is SumA is Sum and B is Group By单项选择题

A

A is Count and B is Group By

B

A is Group By and B is Count

C

A is Group By and B is Sum

D

A is Sum and B is Group By

BlackTom AI · Accepted Answer

Class Name must be grouped and Registration ID must be counted because you want one output row per class and a numeric total of how many registrations fall into each class.

1) Relevant concepts:
- SQL aggregate queries use GROUP BY to collect rows into groups (here, by Class Name) and aggregate functions (COUNT, SUM, etc.) to compute a summary value per group.
- Any non-aggregated field in the SELECT must appear in the GROUP BY clause.
- COUNT counts rows (or non-null values) and is appropriate for “number of registrations.” SUM adds numeric values and is inappropriate for counting rows.

2) Step-by-step reasoning:
- Fields on the design grid: Class Name (from Class table) and Registration ID (from Class Registration table).
- To produce “the number of student registrations for each class,” we must group records by Class Name so each class becomes one group. Therefore A (Total row under Class Name) = Group By.
- For the registrations we need a numeric tally, so we apply an aggregate to Registration ID to count how many registration records exist per class. Therefore B (Total row under Registration ID) = Count.

3) Example intermediate calculation:
- Suppose Class A has Registration IDs 1,2,3 and Class B has IDs 4,5.
- GROUP BY Class Name yields two groups: Class A and Class B.
- COUNT(RegistrationID) for Class A = 3; for Class B = 2.

4) Why correct answer is correct:
- Grouping on Class Name ensures one row per class. Counting Registration ID returns the number of registrations per group, exactly the requested metric.

5) Why other options are incorrect:
- “A is Count and B is Group By” reverses roles—counting Class Name and grouping Registration ID would not produce registrations-per-class.
- “A is Group By and B is Sum” would try to sum registration IDs (nonsensical) instead of counting records.
- “A is Sum and B is Group By” both misplace operations and would not compute counts per class.

Conclusion: A is Group By and B is Count, which matches the provided correct answer.

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