Question

There are two tables: question and answer. In answer I hold user_id and question_id. I want to count how many times each choice is selected.

Below is a working query, but instead of joining the same table 4 times, what is a faster way i.e. joining the answer table only once.

SELECT question.question_id,
    question.correct_choice,
    COUNT(DISTINCT a.user_id) as num_of_a,
    COUNT(DISTINCT b.user_id) as num_of_b,
    COUNT(DISTINCT c.user_id) as num_of_c,
    COUNT(DISTINCT d.user_id) as num_of_d
FROM answer a,
    answer b, 
    answer c, 
    answer d,
    question 
WHERE a.question_id = question.question_id 
    AND b.question_id = question.question_id
    AND c.question_id = question.question_id 
    AND d.question_id = question.question_id 
    AND a.choice = 'A' 
    AND b.choice = 'B'
    AND c.choice = 'C' 
    AND d.choice = 'D'  
GROUP BY question.question_id 
ORDER BY question.question_id asc;

returns

273, D, 5, 2, 8, 39
274, C, 2, 14, 50, 2
277, C, 3, 5, 41, 17
278, C, 16, 9, 34, 9
279, C, 8, 30, 24, 12
280, B, 17, 21, 20, 3
284, C, 2, 3, 19, 1
286, A, 16, 3, 2, 2
287, D, 1, 2, 1, 18
289, B, 3, 18, 2, 2
290, D, 6, 9, 8, 6

Answer 1

This solution only does a single join… additionally, I converted your implicit joins to explicit, and rounded out your GROUP BY:

SELECT 
    q.question_id,
    q.correct_choice,
    COUNT(DISTINCT CASE WHEN a.choice = 'A' THEN a.user_id END) as num_of_a,
    COUNT(DISTINCT CASE WHEN a.choice = 'B' THEN a.user_id END) as num_of_b,
    COUNT(DISTINCT CASE WHEN a.choice = 'C' THEN a.user_id END) as num_of_c,
    COUNT(DISTINCT CASE WHEN a.choice = 'D' THEN a.user_id END) as num_of_d
FROM 
    answer a
    JOIN question q ON a.question_id = q.question_id
GROUP BY q.question_id, q.correct_choice
ORDER BY q.question_id asc;

This works because when the CASE statement doesn’t evaluate to true, it returns NULL, which won’t be included in the COUNT DISTINCT of user Ids.