根据类型计算每月中每一天的行之间的差

Question

我想从MySQL表导出特定月份的数据(基于用户输入)，我不想计算该月份每天的created_at之间的差异。

attendance_id:

• 1 =到达
• 2 = exit
• 10 =在家工作

当用户开始工作时，他的入口记录为1(到达)，当他离开工作时，他的出口记录为2(到达)。用户也可以在家工作，当他们开始工作时，其日志记录为10(在家工作)，当他们停止工作时，其日志记录为2(退出)。表user_attendance

+-----+---------+---------------+----------------+----------------+
| id  | user_id | attendance_id |   created_at   |   updated_at   |
+-----+---------+---------------+----------------+----------------+
|  52 |      45 |             1 | 1.4.2015 6:48  | 1.4.2015 6:48  |
|  53 |      31 |             1 | 1.4.2015 6:52  | 1.4.2015 6:52  |
| 132 |      45 |             2 | 1.4.2015 13:58 | 1.4.2015 13:58 |
| 133 |      31 |             2 | 1.4.2015 14:32 | 1.4.2015 14:32 |
| 159 |      45 |             1 | 2.4.2015 6:49  | 2.4.2015 6:49  |
| 160 |      31 |             1 | 2.4.2015 6:52  | 2.4.2015 6:52  |
| 232 |      31 |             2 | 2.4.2015 15:06 | 2.4.2015 15:06 |
| 233 |      45 |             2 | 2.4.2015 15:09 | 2.4.2015 15:09 |
| 252 |      74 |            10 | 3.4.2015 6:52  | 3.4.2015 6:52  |
| 253 |      74 |             2 | 3.4.2015 15:52 | 3.4.2015 15:52 |
+-----+---------+---------------+----------------+----------------+

SQL:

CREATE TABLE IF NOT EXISTS `user_attendance` (
`id` int(11) NOT NULL,
  `user_id` int(10) unsigned NOT NULL,
  `attendance_id` int(10) NOT NULL,
  `created_at` datetime NOT NULL,
  `updated_at` datetime NOT NULL
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=2836 ;

我的尝试:

SELECT A.user_id, A.created_at,HOUR(TIMEDIFF(B.created_at,A.created_at)) AS timedifference
FROM user_attendance A CROSS JOIN user_attendance B
WHERE B.id IN (SELECT MIN(C.id) FROM user_attendance C WHERE C.id > A.id) and A.user_id=45 and A.attendance_id in(1,2) and MONTH(A.created_at)=5
ORDER BY date(A.created_at) ASC

正如您所看到的，我将其限制为特定的用户和两种类型(到达和退出)，但结果不太好，因为时差为0。

Answer 1

      select  ua.day(a.created_at) as day,ua.user_id, HOUR(timediff(max(created_at),min(created_at))) as TimeDiff
from user_attendance ua
where ua.created_at between '2015-05-01 00:00:00' and '2015-05-31 23:59:59'
group by day(ua.created_at),ua.user_id
order by ua.user_id

Answer 2

在排除错误的情况下，操作顺序总是1-2，而且它在同一天到来和出去，我的解决方案是:

SELECT A.user_id, A.created_at,HOUR(TIMEDIFF(B.created_at,A.created_at))
FROM user_attendance A,  user_attendance B
WHERE
    A.attendance_id in (1,10) AND
    B.attendance_id=2 AND
    A.user_id=45 AND
    B.user_id=A.user_id AND
    DAY(A.created_at)=DAY(B.created_at) AND
    MONTH(A.created_at)=5
ORDER BY date(A.created_at) ASC