Question

我的下表包含消息：

+---------+---------+------------+----------+
| 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，看看每个用户在16:00之前和之后发送了多少条消息。我现在分两步完成：

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，顺便说一下，结果集中没有给出“0”值的行。我只是添加它们以便澄清）

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： - ）

Answer 1

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

在特定时间之前和之后计算每天的消息并显示在一个结果集中

1 个答案: