让每一个问题，许多答对

Question

我有情况下，用户，我需要选择该回答这个问题在SQL 我的表正确看用户的每一个问题数量一样

table 1: questionnaire_question 
id - question_text - thecorrectAnswer 

table 2 :questionnaire_answers 
id - user_id - question_id - user_answer - correct 

正确的列有用户回答值为0或1的正确与否

Answer 1

使用LEFT JOIN，GROUP BY和COUNT

select 
    q.id, 
    count(distinct a.user_id) cnt -- can be replaced with count(*) 
           -- if a user has only at max one correct answer 
           -- for a given question 
from questionnaire_question q 
left join questionnaire_answers a 
on q.id = a.question_id 
where correct = 1 
group by q.id; 

Answer 2

您可以加入表和count一个case表达，检查正确性：

SELECT q.id, cnt 
FROM questionnaire_question q 
JOIN (SELECT question_id, COUNT(CASE correct WHEN 1 THEN 1 END) AS cnt 
     FROM questionnaire_answers 
     GROUP BY question_id) a ON q.id = a.question_id 

Answer 3

使用子查询，如果问题没有正确答案，则返回0

试试这个：

SELECT qq.id, 
    COALESCE((SELECT COUNT(*) 
    FROM questionnaire_answers qa 
    WHERE qa.question_id = qq.question_id 
    AND qa.correct = 1), 0) 
FROM questionnaire_question qq 

Answer 4

select question_id, count(user_id) 
from questionnaire_answers 
where correct = 1 
group by question_id 

没有加入因为您已经通过包含正确的列来非规范化模式。