Question

目前，我一直在寻找最佳查询来制作出勤记录数据库的简历。

这是“ absen”表的当前布局：

id|employee_id|          log        |status
1 | 1         | 2019-03-27 07:58:32 |  in
2 | 1         | 2019-03-27 07:58:38 |  in
3 | 1         | 2019-03-27 17:12:34 |  out
4 | 2         | 2019-03-26 07:45:12 |  in
5 | 2         | 2019-03-26 18:22:54 |  out
6 | 1         | 2019-03-28 07:45:38 |  in
7 | 1         | 2019-03-28 17:10:55 |  out
8 | 2         | 2019-03-27 07:59:02 |  in
9 | 2         | 2019-03-27 18:35:16 |  out

这是目标

employee_id|          in         |        out
    2      | 2019-03-26 07:45:12 | 2019-03-26 18:22:54
    1      | 2019-03-27 07:58:32 | 2019-03-27 17:12:34
    2      | 2019-03-27 07:59:02 | 2019-03-27 18:35:16
    1      | 2019-03-28 07:45:38 | 2019-03-28 17:10:55

（在ASC中为ORDER BY）

我已经尝试过以下查询：

SELECT
    dateng.e_idIN AS "PIN",
    MASUK,
    PULANG
FROM
(
    SELECT
        MIN(log) AS "MASUK",
        employee_id AS e_idIN
    FROM
        absen
    WHERE
        status = "in"
    GROUP BY
        log
) dateng
INNER JOIN(
    SELECT
        MAX(log) AS "PULANG",
        employee_id AS e_idOUT
    FROM
        absen
    WHERE
        status = "out"
    GROUP BY
        log
) pergi
ON
    dateng.e_idIN = pergi.e_idOUT

但结果与预期不符

PIN |         MASUK       |         PULANG
2   | 2019-03-26 07:45:12 | 2019-03-27 18:35:16
2   | 2019-03-26 07:45:12 | 2019-03-26 18:22:54
1   | 2019-03-27 07:58:32 | 2019-03-28 17:10:55
1   | 2019-03-27 07:58:32 | 2019-03-27 17:12:34
1   | 2019-03-27 07:58:38 | 2019-03-28 17:10:55
1   | 2019-03-27 07:58:38 | 2019-03-27 17:12:34
2   | 2019-03-27 07:59:02 | 2019-03-27 18:35:16
2   | 2019-03-27 07:59:02 | 2019-03-26 18:22:54
1   | 2019-03-28 07:45:38 | 2019-03-28 17:10:55
1   | 2019-03-28 07:45:38 | 2019-03-27 17:12:34

此任务有什么线索吗？

Answer 1

我认为这会起作用：

select t.employee_id, min(t.login) login, t.logout  
from (
  select a.employee_id, a.log login, 
  (select min(log) from absen 
   where status = 'out' and employee_id = a.employee_id and log > a.log) logout
  from absen a
  where a.status = 'in'
) t
group by t.employee_id, t.logout  
order by t.login, t.logout

查询移到单独的列

1 个答案: