Question

这就是场景。我有一个MYSQL表用于出勤，其记录如下：

   Id   DateTime             Door   Employee_id
    1   2016-01-01 08:00:00  In     100
    2   2016-01-01 09:00:00  Out    100
    3   2016-01-01 09:11:00  Enter  100
    4   2016-01-01 09:12:00  Exit   100
    5   2016-01-01 09:15:00  In     100
    6   2016-01-01 09:30:00  In     100
    7   2016-01-01 10:00:00  Out    100
    8   2016-01-01 11:00:00  In     100
    9   2016-01-01 12:00:00  In     100
   10   2016-01-01 13:00:00  In     100
   11   2016-01-01 13:30:00  Out    100
   10   2016-01-01 14:00:00  Out    100
   12   2016-01-01 15:00:00  In     100

我希望输出为“最后一个时钟输入”和“最后一个时钟输出”来自＆＃39; In＆＃39;和＆＃39; Out＆＃39;门只如下图所示。如果在最后一个时钟之后没有时钟输出，则忽略时钟输入。如果还有其他门，那么＆＃39; In＆＃39;和＆＃39; Out＆＃39;也忽略了。

Id   Clock In             Clock Out             Employee Id
 1   2016-01-01 08:00:00  2016-01-01 09:00:00   100
 2   2016-01-01 09:30:00  2016-01-01 10:00:00   100
 3   2016-01-01 13:00:00  2016-01-01 14:00:00   100

我已经坚持了很多天了。请帮助我。提前感谢您的帮助。

Answer 1

以下是使用您的数据进行演示的解决方案。

SQL：

-- data
create table attendance(Id int, DateTime datetime, Door char(20), Employee_id int);
INSERT INTO attendance VALUES
(    1,   '2016-01-01 08:00:00',  'In', 100),
(    2,   '2016-01-01 09:00:00',  'Out', 100),
(    3,   '2016-01-01 09:11:00',  'Enter', 100),
(    4,   '2016-01-01 09:12:00',  'Exit', 100),
(    5,   '2016-01-01 09:15:00',  'In', 100),
(    6,   '2016-01-01 09:30:00',  'In', 100),
(    7,   '2016-01-01 10:00:00',  'Out', 100),
(    8,   '2016-01-01 11:00:00',  'In', 100),
(    9,   '2016-01-01 12:00:00',  'In', 100),
(   10,   '2016-01-01 13:00:00',  'In', 100),
(   11,   '2016-01-01 13:30:00',  'Out', 100),
(   10,   '2016-01-01 14:00:00',  'Out', 100),
(   12,   '2016-01-01 15:00:00',  'In', 100);
SELECT * FROM attendance;

-- SQL needed:
SELECT  @id:=@id+1 Id,
        MAX(IF(Door = 'In', DateTime, NULL)) `Check In`,
        MAX(IF(Door = 'Out', DateTime, NULL)) `Check Out`,
        Employee_id
FROM (SELECT *, 
            CASE WHEN (Door != 'Out' AND @last_door = 'Out')
                 THEN @group_num:=@group_num+1
                 ELSE @group_num END door_group,
            @last_door:=Door
        FROM attendance JOIN (SELECT @group_num:=1, @last_door := NULL) a
      ) t JOIN (SELECT @id:=0 ) b
GROUP BY t.door_group
HAVING SUM(Door = 'In') > 0 AND SUM(Door = 'Out') > 0;

输出：

mysql> SELECT * FROM attendance;
:+------+---------------------+-------+-------------+
| Id   | DateTime            | Door  | Employee_id |
+------+---------------------+-------+-------------+
|    1 | 2016-01-01 08:00:00 | In    |         100 |
|    2 | 2016-01-01 09:00:00 | Out   |         100 |
|    3 | 2016-01-01 09:11:00 | Enter |         100 |
|    4 | 2016-01-01 09:12:00 | Exit  |         100 |
|    5 | 2016-01-01 09:15:00 | In    |         100 |
|    6 | 2016-01-01 09:30:00 | In    |         100 |
|    7 | 2016-01-01 10:00:00 | Out   |         100 |
|    8 | 2016-01-01 11:00:00 | In    |         100 |
|    9 | 2016-01-01 12:00:00 | In    |         100 |
|   10 | 2016-01-01 13:00:00 | In    |         100 |
|   11 | 2016-01-01 13:30:00 | Out   |         100 |
|   10 | 2016-01-01 14:00:00 | Out   |         100 |
|   12 | 2016-01-01 15:00:00 | In    |         100 |
+------+---------------------+-------+-------------+
13 rows in set (0.00 sec)

mysql>
mysql> -- SQL needed:
mysql> SELECT  @id:=@id+1 Id,
    ->         MAX(IF(Door = 'In', DateTime, NULL)) `Check In`,
    ->         MAX(IF(Door = 'Out', DateTime, NULL)) `Check Out`,
    ->         Employee_id
    -> FROM (SELECT *,
    ->             CASE WHEN (Door != 'Out' AND @last_door = 'Out')
    ->                  THEN @group_num:=@group_num+1
    ->                  ELSE @group_num END door_group,
    ->             @last_door:=Door
    ->         FROM attendance JOIN (SELECT @group_num:=1, @last_door := NULL) a
    ->       ) t JOIN (SELECT @id:=0 ) b
    -> GROUP BY t.door_group
    -> HAVING SUM(Door = 'In') > 0 AND SUM(Door = 'Out') > 0;
+------+---------------------+---------------------+-------------+
| Id   | Check In            | Check Out           | Employee_id |
+------+---------------------+---------------------+-------------+
|    1 | 2016-01-01 08:00:00 | 2016-01-01 09:00:00 |         100 |
|    2 | 2016-01-01 09:30:00 | 2016-01-01 10:00:00 |         100 |
|    3 | 2016-01-01 13:00:00 | 2016-01-01 14:00:00 |         100 |
+------+---------------------+---------------------+-------------+
3 rows in set (0.00 sec)

Answer 2

试试此代码

Select @x := -1;
Select @pDate := NOW();

SELECT all2.link, all2.prevD as 'Clock In', all2.DateTime as 'Clock Out'
FROM
 (SELECT
    DateTime, id, Door, @x:=@x AS prev,
    if(@x > 0, @x, -1) as link,
    @x:=IF(door = 'In', id, if(door = 'Out', id, -1)) AS cure,
    @pDate as prevD,
    @pDate:=IF(door = 'In', DateTime, NULL) AS pDate,
    @x:=IF(door = 'Out', -1, @x) as reseter
FROM
    test.doors) as all2
WHERE all2.link != -1 AND all2.Door = 'Out'

实际上有一个错误是： “我无法识别最后Out州，所以我希望你能尝试解决它，我没有足够的时间。

祝福

获得最后进入和最后出来的特殊门只有MYSQL

2 个答案: