SQL Select 一个月的考勤表
SQL Select Statement for Time and attendance for a month
有人可以帮忙解决这个问题吗?我们的考勤系统生成以下数据:
Empid Department Timestamp Read_ID
3221 IT 2017-01-29 11:12:00.000 1
5565 IT 2017-01-29 12:28:06.000 1
5565 IT 2017-01-29 12:28:07.000 1
3221 IT 2017-01-29 13:12:00.000 2
5565 IT 2017-01-29 13:28:06.000 2
3221 IT 2017-01-30 07:42:15.000 1
3221 IT 2017-01-30 16:16:15.000 2
3221 IT 2017-01-31 09:05:00.000 1
3221 IT 2017-01-31 11:05:00.000 2
3221 IT 2017-01-31 13:20:00.000 1
3221 IT 2017-01-31 16:10:00.000 2
其中 Read_ID 值为:
1
= 条目
2
= 退出
我正在寻找 SQL 在 MS SQL 2014 上对 运行 的查询,该查询汇总了每个员工每月的出勤时间,例如
Empid Department Year Month TotalHours
3221 IT 2017 1 15:24
5565 IT 2017 1 01:00
试试这个。我不确定哪种时间格式可以满足您的系统要求,所以我将两者都放在了:
SELECT * INTO #Tbl3 FROM (VALUES
(3221,'IT','2017-01-29 11:12:00.000',1),
(5565,'IT','2017-01-29 12:28:06.000',1),
(5565,'IT','2017-01-29 12:28:07.000',1),
(3221,'IT','2017-01-29 13:12:00.000',2),
(5565,'IT','2017-01-29 13:28:06.000',2),
(3221,'IT','2017-01-30 07:42:15.000',1),
(3221,'IT','2017-01-30 16:16:15.000',2),
(3221,'IT','2017-01-31 09:05:00.000',1),
(3221,'IT','2017-01-31 11:05:00.000',2),
(3221,'IT','2017-01-31 13:20:00.000',1),
(3221,'IT','2017-01-31 16:10:00.000',2))
x (Empid,Department,Timestamp,Read_ID)
;With cte as (
SELECT t1.Empid, t1.Department
, [Year] = Year(t1.Timestamp)
, [Month] = Month(t1.Timestamp)
, Seconds = SUM(DATEDIFF(second, t1.Timestamp, t2.Timestamp))
FROM #Tbl3 as t1
OUTER APPLY (
SELECT Timestamp = MIN(t.Timestamp)
FROM #Tbl3 as t
WHERE t.Department = t1.Department and t.Empid = t1.Empid
and t.Timestamp > t1.Timestamp and t.Read_ID = 2
) as t2
WHERE t1.Read_ID = 1
GROUP BY t1.Empid, t1.Department, Year(t1.Timestamp), Month(t1.Timestamp))
SELECT *, TotalHours = Seconds / 3600., TotalTime =
RIGHT('0'+CAST(Seconds / 3600 as VARCHAR),2) + ':' +
RIGHT('0'+CAST((Seconds % 3600) / 60 as VARCHAR),2) + ':' +
RIGHT('0'+CAST(Seconds % 60 as VARCHAR),2)
FROM cte;
此查询应该会为您提供所需的结果。它的工作原理是选择每个条目,并将其与同一员工的下一个退出结合起来(没有进一步退出的条目将被忽略):这为我们提供了该员工轮班的持续时间。然后汇总结果并计算每组的班次持续时间。
SELECT
t1.empid,
t1.department,
YEAR(t1.timestamp) Year,
MONTH(t1.timestamp) Month,
CONVERT(
varchar(12),
DATEADD(minute, SUM(DATEDIFF(minute, t1.timestamp, t2.timestamp)), 0),
114
) TotalHours
FROM
mytable t1
INNER JOIN mytable t2
ON t1.empid = t2.empid
AND t2.read_id = 2
AND t2.timestamp = (
SELECT MIN(timestamp)
FROM mytable
WHERE
read_id = 2
AND empid = t2.empid
AND timestamp > t1.timestamp
)
WHERE
t1.read_id = 1
GROUP BY t1.empid, t1.department, YEAR(t1.timestamp), MONTH(t1.timestamp)
ORDER BY 1, 2, 3, 4
Returns :
empid | department | Year | Month | TotalHours
----: | :--------- | ---: | ----: | :-----------
3221 | IT | 2017 | 1 | 15:24:00:000
5565 | IT | 2017 | 1 | 02:00:00:000
DB Fiddle demo on SQL Server 2014
但是,存在一种极端情况,员工进入两次然后存在(这发生在您的数据中,员工 5565
在 29/01/2017 12:28:06
和 在 29/01/2017 12:28:07
,然后在 29/01/2017 13:28:06
退出。上面的查询将考虑到两个重叠的条目并将它们映射到同一个出口,导致这一小时的工作被计算两次。
虽然这符合您的预期结果,但这是您真正想要的吗?这是一个替代查询,如果连续出现多个相同的员工条目,则只考虑最新的一个:
SELECT
t1.empid,
t1.department,
YEAR(t1.timestamp) Year,
MONTH(t1.timestamp) Month,
CONVERT(
varchar(12),
DATEADD(minute, SUM(DATEDIFF(minute, t1.timestamp, t2.timestamp)), 0),
114
) TotalHours
FROM
mytable t1
INNER JOIN mytable t2
ON t1.empid = t2.empid
AND t2.read_id = 2
AND t2.timestamp = (
SELECT MIN(timestamp)
FROM mytable
WHERE
read_id = 2
AND empid = t2.empid
AND timestamp > t1.timestamp
)
WHERE
t1.read_id = 1
AND NOT EXISTS (
SELECT 1
FROM mytable
WHERE
read_id = 1
AND empid = t1.empid
AND timestamp > t1.timestamp
AND timestamp < t2.timestamp
)
GROUP BY t1.empid, t1.department, YEAR(t1.timestamp), MONTH(t1.timestamp)
ORDER BY 1, 2, 3, 4
Returns :
empid | department | Year | Month | TotalHours
----: | :--------- | ---: | ----: | :-----------
3221 | IT | 2017 | 1 | 15:24:00:000
5565 | IT | 2017 | 1 | 01:00:00:000
有人可以帮忙解决这个问题吗?我们的考勤系统生成以下数据:
Empid Department Timestamp Read_ID
3221 IT 2017-01-29 11:12:00.000 1
5565 IT 2017-01-29 12:28:06.000 1
5565 IT 2017-01-29 12:28:07.000 1
3221 IT 2017-01-29 13:12:00.000 2
5565 IT 2017-01-29 13:28:06.000 2
3221 IT 2017-01-30 07:42:15.000 1
3221 IT 2017-01-30 16:16:15.000 2
3221 IT 2017-01-31 09:05:00.000 1
3221 IT 2017-01-31 11:05:00.000 2
3221 IT 2017-01-31 13:20:00.000 1
3221 IT 2017-01-31 16:10:00.000 2
其中 Read_ID 值为:
1
= 条目2
= 退出
我正在寻找 SQL 在 MS SQL 2014 上对 运行 的查询,该查询汇总了每个员工每月的出勤时间,例如
Empid Department Year Month TotalHours
3221 IT 2017 1 15:24
5565 IT 2017 1 01:00
试试这个。我不确定哪种时间格式可以满足您的系统要求,所以我将两者都放在了:
SELECT * INTO #Tbl3 FROM (VALUES
(3221,'IT','2017-01-29 11:12:00.000',1),
(5565,'IT','2017-01-29 12:28:06.000',1),
(5565,'IT','2017-01-29 12:28:07.000',1),
(3221,'IT','2017-01-29 13:12:00.000',2),
(5565,'IT','2017-01-29 13:28:06.000',2),
(3221,'IT','2017-01-30 07:42:15.000',1),
(3221,'IT','2017-01-30 16:16:15.000',2),
(3221,'IT','2017-01-31 09:05:00.000',1),
(3221,'IT','2017-01-31 11:05:00.000',2),
(3221,'IT','2017-01-31 13:20:00.000',1),
(3221,'IT','2017-01-31 16:10:00.000',2))
x (Empid,Department,Timestamp,Read_ID)
;With cte as (
SELECT t1.Empid, t1.Department
, [Year] = Year(t1.Timestamp)
, [Month] = Month(t1.Timestamp)
, Seconds = SUM(DATEDIFF(second, t1.Timestamp, t2.Timestamp))
FROM #Tbl3 as t1
OUTER APPLY (
SELECT Timestamp = MIN(t.Timestamp)
FROM #Tbl3 as t
WHERE t.Department = t1.Department and t.Empid = t1.Empid
and t.Timestamp > t1.Timestamp and t.Read_ID = 2
) as t2
WHERE t1.Read_ID = 1
GROUP BY t1.Empid, t1.Department, Year(t1.Timestamp), Month(t1.Timestamp))
SELECT *, TotalHours = Seconds / 3600., TotalTime =
RIGHT('0'+CAST(Seconds / 3600 as VARCHAR),2) + ':' +
RIGHT('0'+CAST((Seconds % 3600) / 60 as VARCHAR),2) + ':' +
RIGHT('0'+CAST(Seconds % 60 as VARCHAR),2)
FROM cte;
此查询应该会为您提供所需的结果。它的工作原理是选择每个条目,并将其与同一员工的下一个退出结合起来(没有进一步退出的条目将被忽略):这为我们提供了该员工轮班的持续时间。然后汇总结果并计算每组的班次持续时间。
SELECT
t1.empid,
t1.department,
YEAR(t1.timestamp) Year,
MONTH(t1.timestamp) Month,
CONVERT(
varchar(12),
DATEADD(minute, SUM(DATEDIFF(minute, t1.timestamp, t2.timestamp)), 0),
114
) TotalHours
FROM
mytable t1
INNER JOIN mytable t2
ON t1.empid = t2.empid
AND t2.read_id = 2
AND t2.timestamp = (
SELECT MIN(timestamp)
FROM mytable
WHERE
read_id = 2
AND empid = t2.empid
AND timestamp > t1.timestamp
)
WHERE
t1.read_id = 1
GROUP BY t1.empid, t1.department, YEAR(t1.timestamp), MONTH(t1.timestamp)
ORDER BY 1, 2, 3, 4
Returns :
empid | department | Year | Month | TotalHours ----: | :--------- | ---: | ----: | :----------- 3221 | IT | 2017 | 1 | 15:24:00:000 5565 | IT | 2017 | 1 | 02:00:00:000
DB Fiddle demo on SQL Server 2014
但是,存在一种极端情况,员工进入两次然后存在(这发生在您的数据中,员工 5565
在 29/01/2017 12:28:06
和 在 29/01/2017 12:28:07
,然后在 29/01/2017 13:28:06
退出。上面的查询将考虑到两个重叠的条目并将它们映射到同一个出口,导致这一小时的工作被计算两次。
虽然这符合您的预期结果,但这是您真正想要的吗?这是一个替代查询,如果连续出现多个相同的员工条目,则只考虑最新的一个:
SELECT
t1.empid,
t1.department,
YEAR(t1.timestamp) Year,
MONTH(t1.timestamp) Month,
CONVERT(
varchar(12),
DATEADD(minute, SUM(DATEDIFF(minute, t1.timestamp, t2.timestamp)), 0),
114
) TotalHours
FROM
mytable t1
INNER JOIN mytable t2
ON t1.empid = t2.empid
AND t2.read_id = 2
AND t2.timestamp = (
SELECT MIN(timestamp)
FROM mytable
WHERE
read_id = 2
AND empid = t2.empid
AND timestamp > t1.timestamp
)
WHERE
t1.read_id = 1
AND NOT EXISTS (
SELECT 1
FROM mytable
WHERE
read_id = 1
AND empid = t1.empid
AND timestamp > t1.timestamp
AND timestamp < t2.timestamp
)
GROUP BY t1.empid, t1.department, YEAR(t1.timestamp), MONTH(t1.timestamp)
ORDER BY 1, 2, 3, 4
Returns :
empid | department | Year | Month | TotalHours ----: | :--------- | ---: | ----: | :----------- 3221 | IT | 2017 | 1 | 15:24:00:000 5565 | IT | 2017 | 1 | 01:00:00:000