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mysql - SQL SERVER (TSQL) 逆 DateDiff 的 SUM(SQL 或存储过程)

转载 作者:太空宇宙 更新时间:2023-11-03 12:01:20 27 4
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我有一个棘手的 SQL 需要创建,有这个表:

+-----------------------+
| Employee Login Logout |
+-----------------------+
| 1 08:30 08:50 |
| 1 09:00 10:00 |
| 1 10:20 11:00 |
+-----------------------+

我需要一个 sql 来总结所有的休息时间:

08:50 -> 09:00 = 10 minutes +
10:00 -> 10:20 = 20 minutes =
30 minutes in total

所以我需要以某种方式检查每条记录并使用 datediff 方法 sum 记录之间的中断持续时间。

有什么简单的方法可以使用 SQLStored Procedure 来实现吗?

最佳答案

正如@thebjorn 所评论的那样,您可以首先获得第一次登录和最后一次注销(轮类)之间的差异,然后减去每个登录-注销对的差异的SUM(实际时间正在工作)。结果将是您不工作的总时间或您的休息时间。

;WITH SampleData(Employee, Login, Logout) AS(
SELECT 1, CAST('08:30' AS TIME), CAST('08:50' AS TIME) UNION ALL
SELECT 1, CAST('09:00' AS TIME), CAST('10:00' AS TIME) UNION ALL
SELECT 1, CAST('10:20' AS TIME), CAST('11:00' AS TIME)
)
SELECT
Employee,
TotalBreakTime = DATEDIFF(MINUTE, MIN(Login), MAX(Logout)) - SUM(DATEDIFF(MINUTE, Login, Logout))
FROM SampleData
GROUP BY Employee

使用LAG函数

;WITH SampleData(Employee, Login, Logout) AS(
SELECT 1, CAST('08:30' AS TIME), CAST('08:50' AS TIME) UNION ALL
SELECT 1, CAST('09:00' AS TIME), CAST('10:00' AS TIME) UNION ALL
SELECT 1, CAST('10:20' AS TIME), CAST('11:00' AS TIME) UNION ALL
SELECT 2, CAST('10:00' AS TIME), CAST('10:30' AS TIME) UNION ALL
SELECT 2, CAST('10:45' AS TIME), CAST('11:30' AS TIME)
)
,CteBreakTime AS(
SELECT *,
BreakTime = DATEDIFF(MINUTE, LAG(Logout) OVER(PARTITION BY Employee ORDER BY Login), Login)
FROM SampleData
)
SELECT
Employee,
TotalBreakTime = SUM(BreakTime)
FROM CteBreakTime
GROUP BY Employee

关于mysql - SQL SERVER (TSQL) 逆 DateDiff 的 SUM(SQL 或存储过程),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29091671/

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