Mysql 具有时差的员工休息时间历史记录

Question

我有带有 inoutmode 标志的员工生物识别日志数据。我正在尝试获取详细的休息时间列表和时差。

inoutmode 4 次突破，5 次闯入。

INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (105, '2019-09-19', '14:00:13', 4);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (105, '2019-09-19', '16:07:08', 4);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (105, '2019-09-19', '16:07:18', 5);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (235, '2019-09-19', '15:44:26', 4);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (235, '2019-09-19', '16:37:58', 4);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (235, '2019-09-19', '20:01:11', 5);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (235, '2019-09-19', '20:01:25', 5);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (235, '2019-09-19', '20:30:29', 4);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (326, '2019-09-19', '15:58:30', 4);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (326, '2019-09-19', '19:34:09', 5);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (327, '2019-09-19', '15:44:19', 5);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (327, '2019-09-19', '15:55:37', 4);
INSERT INTO `tbl_downloadentry` (`EmpMachineID`, `shift_date`, `AttenTime`, `InOutMode`) VALUES (327, '2019-09-19', '19:59:38', 4);

这是我的期望输出

| EmpMachineID | attendance_date | break_out  | break_in  | Diff      |
|--------------|-----------------|------------|-----------|-----------|
| 235          | 2019-09-19      | 15:44:26   |           |           |
|              | 2019-09-19      | 16:37:58   |           |           |
|              | 2019-09-19      |            | 20:01:11  |           |
|              | 2019-09-19      | 20:30:29   | 20:01:25  |    29:04  |
| 326          | 2019-09-19      | 19:34:09   | 15:58:30  | 03:35:39  |

我尽力实现了输出。以下是我尝试的查询：

SELECT l2.empmachineid,
       l2.shift_date,
       l2.attentime,
       l2.inoutmode
FROM   tbl_downloadentry AS l2
WHERE  l2.inoutmode IN ( 5, 4 )
       AND l2.shift_date = "2019-09-19"
ORDER  BY l2.empmachineid,
          l2.shift_date,
          l2.attentime ASC

我的MySQL版本=10.3.17-MariaDB-1-log

SELECT l2.EmpMachineID, l2.shift_date, l2.InOutMode, 
       case when l2.InOutMode=5 then l2.AttenTime END AS BreakOut, 
       case when l2.InOutMode=4 then l2.AttenTime END AS BreakIn
FROM tbl_downloadentry AS l2
WHERE l2.InOutMode IN (5, 4) AND l2.shift_date="2019-09-19"
ORDER BY l2.EmpMachineID, l2.shift_date, l2.AttenTime ASC

部分我使用案例条件到达了记录，但 4 和 5 inoutmode 应该在单行中以计算时差。任何想法...

如有任何建议，我们将不胜感激。

Answer 1

如果我理解正确的话，你想将“4”条记录与“5”条记录相关联。你的结果集好像和你的样本数据关系不大，所以很难理解。

以下方法将每个 4/5 与最近的一个相关联。它通过分配一个分组来做到这一点，对每个“4”进行向上计数，对每个“5”进行向下计数。

然而，这并不是完整的解决方案，因为连续“4"s and "5”系列可能会有多次起伏。所以，它分配了一个辅助分组，将其分解。

SELECT EmpMachineID, shift_date,
       MAX(CASE WHEN InOutMode = 4 THEN AttenTime END),
       MAX(CASE WHEN InOutMode = 5 THEN AttenTime END)
FROM (SELECT dl.*,
             SUM(InOutMode = group_first_InOutMode) OVER (PARTITION BY dl.EmpMachineID, dl.shift_date, dl.grouping ORDER BY dl.AttenTime) as secondary_grouping
      FROM (SELECT dl.*,
                   FIRST_VALUE(InOutMode) OVER (PARTITION BY dl.EmpMachineID, dl.shift_date, dl.grouping ORDER BY dl.AttenTime) as group_first_InOutMode
            FROM (SELECT dl.*,
                         SUM( CASE WHEN InOutMode = 4 THEN 1 WHEN InOutMode = 5 THEN -1 END) OVER
                              (PARTITION BY dl.EmpMachineID, dl.shift_date
                               ORDER BY dl.AttenTime
                              ) -
                              (CASE WHEN InOutMode = 5 THEN -1 ELSE 0 END) as grouping  -- subtract out "5"s on current row
                 FROM tbl_downloadentry dl
                 WHERE dl.InOutMode IN (5, 4) AND dl.shift_date = '2019-09-19'
                ) dl
           ) dl
     ) dl
GROUP BY EmpMachineID, shift_date, grouping, secondary_grouping;

Here 是一个 db<>fiddle.

Answer 2

这是一种利用 MariaDB 10.2+ 和 MySQL 8+ 中可用的 LEAD() and LAG() Window 函数的方法。

当一个特定的行InOutMode模式是4时，这意味着它是break_out时间。现在，我们使用 LAG() 函数获取该特定 EmpID 的直接 previous 行。排序是根据时间定义的。因此，如果紧邻的前一行的 InOutMode 模式是 5，这意味着我们有一个相应的 break_in 时间用于此 break_out 时间，否则 null.
InOutMode 模式为 5 的行遵循类似的过程。这次唯一不同的是我们使用了 LEAD() 函数；因为我们需要立即获取 next 行，并检查它是否是 break_out。
现在，我们只需要将此结果集用作 Derived Table and DISTINCT it out (because we will have duplicate rows for every case where there are break_in and break_out together). Also, in the outer query we can calculate the time difference using TimeDiff() 功能。

为了演示目的，为 EmpID = 235 完成了以下查询：

SELECT 
  DISTINCT 
    dt.*, 
    TIMEDIFF(dt.break_out, dt.break_in) AS diff 
FROM 
(
SELECT  
  EmpMachineID, 
  shift_date, 
  CASE InOutMode 
    WHEN 4 THEN AttenTime -- this is break_out row
    WHEN 5 THEN -- this is break_in row, find the break_out if exists
      CASE 
        WHEN LEAD(InOutMode) OVER w = 4 
        THEN LEAD(AttenTime) OVER w 
      END       
  END AS break_out, 
  CASE InOutMode 
    WHEN 5 THEN AttenTime -- this is break_in row
    WHEN 4 THEN   -- this is break_out row, find the break_in if exists
      CASE 
        WHEN LAG(InOutMode) OVER w = 5 
        THEN LAG(AttenTime) OVER w 
      END  
  END AS break_in
FROM tbl_downloadentry 
WHERE EmpMachineID = 235 
AND InOutMode IN (4,5) 
AND shift_date = '2019-09-19' 
WINDOW w AS (PARTITION BY EmpMachineID
             ORDER BY AttenTime ASC)
) AS dt;

结果

| EmpMachineID | shift_date | break_out | break_in | diff     |
| ------------ | ---------- | --------- | -------- | -------- |
| 235          | 2019-09-19 | 15:44:26  |          |          |
| 235          | 2019-09-19 | 16:37:58  |          |          |
| 235          | 2019-09-19 |           | 20:01:11 |          |
| 235          | 2019-09-19 | 20:30:29  | 20:01:25 | 00:29:04 |

View on DB Fiddle

Mysql 具有时差的员工休息时间历史记录

Mysql Employee break time history with time difference

mysql

sql

mariadb

mariadb-10.3