Question

我有一张这样的表：

// notifications
+----+--------+-----------+---------+--------------------+
| id | money  | post_id   | user_id | belongs_to_user_id |
+----+--------+-----------+---------+--------------------+
| 1  | 5      | 1         | 123     | 101                |
| 2  | 10     | 2         | 123     | 101                |
| 3  | -2     | 4         | 456     | 101                |
| 5  | -2     | 2         | 456     | 101                |
| 6  | -2     | 3         | 123     | 101                |
| 7  | 5      | 4         | 789     | 101                |
| 8  | 10     | 4         | 789     | 101                |
+----+--------+-----------+---------+--------------------+

这是我的疑问：

SELECT * FROM notifications
WHERE belongs_to_user_id = 101
GROUP BY post_id, user_id
ORDER BY id DESC
LIMIT 3

当前输出应该是这样的：

+----+--------+-----------+---------+--------------------+
| 5  | -2     | 2         | 456     | 101                |
| 6  | -2     | 3         | 123     | 101                |
| 8  | 10     | 4         | 789     | 101                |
+----+--------+-----------+---------+--------------------+

第七行被分组，我们无法在结果中看到它。这正是问题所在。以下是预期结果：

+----+--------+-----------+---------+--------------------+
| 5  | -2     | 2         | 456     | 101                |
| 6  | -2     | 3         | 123     | 101                |
| 7  | 5      | 4         | 789     | 101                |
| 8  | 10     | 4         | 789     | 101                |
+----+--------+-----------+---------+--------------------+

如果我删除GROUP BY，则第五个将被省略。所以这是逻辑：

我想要最后三行（无论分组）。换句话说，嗯......很难说，我想选择分组行（但不计入LIMIT）。

知道我该怎么做？

Answer 1

请尝试此查询。它将获取最后三个“组”，然后提取这些组的所有行（使用连接）：

更新：在子查询中使用DISTINCT的相同查询：

SELECT t.*
  FROM notifications t
 INNER JOIN (SELECT s.post_id, s.user_id
               FROM notifications s
              WHERE belongs_to_user_id = 101
              GROUP BY post_id, user_id
              ORDER BY post_id DESC, user_id DESC
              LIMIT 3) u
    ON u.post_id = t.post_id
   AND u.user_id = t.user_id
 WHERE t.belongs_to_user_id = 101
 ORDER BY t.id

Answer 2

请试试这个

SELECT * FROM notifications WHERE belongs_to_user_id = 101 GROUP BY id, money ORDER BY id DESC

Answer 3

所以，如果我没错，你想要相关subquery

select * from table t
where belongs_to_user_id = 101 and 
      user_id = (select max(user_id) from table where post_id = t.post_id)

此外，您可以添加limit子句来限制所需的记录。

Answer 4

按组显示逗号分隔的ID

SELECT GROUP_CONCAT（id），post_id FROM notifications WHERE belongs_to_user_id = 101 GROUP BY post_id，user_id ORDER BY id DESC 限制3

如何获取分组行的ID？

4 个答案: