Question

我有user_answers表：

id | quiz_id | user_id | question_id | choosen_answer_id |
1  |    1    |    2    |      1      |        5          |  
2  |    1    |    2    |      2      |        6          |  
3  |    1    |    2    |      3      |        7          |  
4  |    1    |    3    |      1      |        5          |  
5  |    1    |    3    |      2      |        8          |  
6  |    1    |    3    |      3      |        9          |

我想cont与{{1}的choosen_answer_id用户相比，{{1}的用户有多少正确答案（相同question_id的相同user_id = 3） }}

在此示例中，我应该收到user_id = 2，因为具有quiz_id的用户仅对问题为id_id = 1的问题回答正确（对于user_id = 2，question_id = 1 selectn_answer_id = 5，对于user_id = 3，question_id = 1 choicen_answer_id = 5`）。

谢谢

Answer 1

我了解您愿意考虑两个用户给出相同答案的问题数量。您可以按如下方式自动加入表格：

SELECT COUNT(*) AS number_of_identical_answers
FROM user_answers ua2
INNER JOIN user_answers ua3 
    ON  ua3.user_id           = 3 
    AND ua3.quiz_id           = ua2.quiz_id 
    AND ua3.question_id       = ua2.question_id 
    AND ua3.choosen_answer_id = ua2.choosen_answer_id
WHERE ua2.user_id = 2

如果要显示合并的记录而不是对它们进行计数，则可以将COUNT(*)更改为表ua2（用户2的答案）和ua3（答案的表）的列列表用户3）。

Answer 2

select user_id, count(*)
from table1 t1
join (select *from table1 where user_id = 2) t2
on t2.question_id = t1.question_id
where t2.choosen_answer_id  = t1.choosen_answer_id 
group by t1.user_id

MySQL计数多少个答案是正确的

2 个答案: