Question

我有3张桌子：

申请（有很多票）
投票（属于申请和问题）
问题（有很多选票）

我需要为每个问题获得每个申请的投票数。

所以，我的尝试是：

SELECT applications.id, COUNT(votes.id), votes.question_id
FROM applications
LEFT OUTER JOIN votes ON (votes.application_id = application.id)
GROUP BY votes.question_id

但是，它仅显示单个应用程序的数据，因此我认为我的查询格式错误：

+----+-----------------+-------------+
| id | COUNT(votes.id) | question_id |
+----+-----------------+-------------+
|  1 |            1185 |           1 |
|  1 |            1170 |           2 |
|  1 |            1209 |           3 |
|  1 |            1230 |           4 |
|  1 |            1213 |           5 |
+----+-----------------+-------------+

我需要什么：

+----+-----------------+-------------+
| id | COUNT(votes.id) | question_id |
+----+-----------------+-------------+
|  1 |            1185 |           1 |
|  1 |            1170 |           2 |
|  1 |            1209 |           3 |
|  1 |            1230 |           4 |
|  1 |            1213 |           5 |
|  2 |            null |           1 |
|  2 |            50   |           2 |
|  2 |            333  |           3 |
|  2 |            1230 |           4 |
|  2 |            1213 |           5 |
|  3 |            null |           1 |
|  3 |            50   |           2 |
|  3 |            333  |           3 |
|  3 |            null |           4 |
|  3 |            5555 |           5 |
+----+-----------------+-------------+

Answer 1

group by条款缺少applications.id。

SELECT applications.id, COUNT(votes.id), votes.question_id
FROM applications
LEFT OUTER JOIN votes ON votes.application_id = application.id
group by applications.id, votes.question_id

Answer 2

您应该按applications.id以及questions.id：

进行分组

SELECT a.id, COUNT(votes.id), votes.question_id
FROM applications a LEFT OUTER JOIN
     votes v
     ON v.application_id = a.id
GROUP BY a.id, v.question_id;

然而，这不会产生你想要的。您似乎想要申请的所有问题，无论是否有任何投票。如果是这样，这可能就是你想要的：

SELECT a.id, q.question_id, COUNT(v.application_id)
FROM applications a CROSS JOIN
     (SELECT DISTINCT question_id FROM votes) q LEFT JOIN
     votes v
     ON v.application_id = a.id and v.question_id = q.question_id
GROPU BY a.id, q.question_id;

使用mysql中的group子句连接count表

2 个答案: