4

我有下表的消息：

+---------+---------+------------+----------+
| msg_id  | user_id | _date      | _time    |
+-------------------+------------+----------+
|   1     | 1       | 2011-01-22 | 06:23:11 |
|   2     | 1       | 2011-01-23 | 16:17:03 |
|   3     | 1       | 2011-01-23 | 17:05:45 |
|   4     | 2       | 2011-01-22 | 23:58:13 |
|   5     | 2       | 2011-01-23 | 23:59:32 |
|   6     | 3       | 2011-01-22 | 13:45:00 |
|   7     | 3       | 2011-01-23 | 13:22:34 |
|   8     | 3       | 2011-01-23 | 18:22:34 |
+---------+---------+------------+----------+

我想要的是每天查看每个用户在 16:00 之前和之后发送了多少条消息。我现在分两步执行此操作：

SELECT user_id, _date, COUNT(msg_id) AS cnt 
FROM messages WHERE _time <= '16:00'
GROUP BY user_id, _date ORDER BY user_id, _date ASC

user_id _date       cnt
-----------------------------
1       2011-01-22  1
1       2011-01-23  0
2       2011-01-22  0
2       2011-01-23  0
3       2011-01-22  1
3       2011-01-23  1

SELECT user_id, _date, COUNT(msg_id) AS cnt 
FROM messages WHERE _time > '16:00'
GROUP BY user_id, _date ORDER BY user_id, _date ASC

user_id _date       cnt
-----------------------------
1       2011-01-22  0
1       2011-01-23  2
2       2011-01-22  1
2       2011-01-23  1
3       2011-01-22  0
3       2011-01-23  1

（实际上，顺便说一句，结果集中没有给出具有“0”值的行。我只是为了澄清而添加了它们）

我想将这两个输出合并为一个：

user_id _date       before16  after16
-------------------------------------
1       2011-01-22  1         0
1       2011-01-23  0         2
2       2011-01-22  0         1
2       2011-01-23  0         1
3       2011-01-22  1         0
3       2011-01-23  1         1

但是，我不知道如何编写此查询。如果你这样做，你的帮助将不胜感激:-)

4

1 回答 1

4

试试这个：

SELECT 
    user_id, 
    _date, 
    SUM(_time <= '16:00') AS before16, 
    SUM(_time > '16:00') AS after16 
FROM messages 
GROUP BY user_id, _date
ORDER BY user_id, _date ASC
于 2012-04-11T18:47:28.333 回答