基本上我想知道，当商店开放和关闭时，有多少员工，详情如下
EmployeeID | IN and OUT Time | Type | InOut ID | Comments |
---------- | ----------------------- |------------| ---------|-------------|
12961 2017-04-24 08:07:00.000 Punch In 1 Store Open
12680 2017-04-24 08:07:00.000 Punch In 2
12662 2017-04-24 08:07:00.000 Punch In 3
12683 2017-04-24 08:27:00.000 Punch In 4
12864 2017-04-24 08:42:00.000 Punch In 5
12681 2017-04-24 10:03:00.000 Punch In 6
-1 2017-04-24 13:33:00.000 Punch In 7
12662 2017-04-24 18:00:00.000 Punch Out 8
12683 2017-04-24 18:00:00.000 Punch Out 9
12864 2017-04-24 18:35:00.000 Punch Out 10
12681 2017-04-24 22:00:00.000 Punch Out 11
12960 2017-04-24 22:00:00.000 Punch Out 12
12959 2017-04-24 22:00:00.000 Punch Out 13
-1 2017-04-24 22:00:00.000 Punch Out 14 Store Close
结果：
Header | Header
-------------------- | ------
No. Of Emp At Open | 3
No. Of Emp At Close | 4
答案 0 :(得分：0)
您可以使用以下查询来查找员工是否仍在办公室。然后，您可以根据此查询的输出计算办公室和办公室外的员工编号。
SELECT in.EmployeeID ,
CASE WHEN out.EmployeeID IS NULL
THEN 1
ELSE 0
END AS Still_In_Office
FROM Employee in
LEFT OUTER JOIN Employee out ON in.EmployeeID = out.EmployeeID
AND out.Type = "Punch Out"
WHERE in.Type = "Punch In"
答案 1 :(得分：0)
select 'No. Of Emp At Open',count([IN and OUT Time]) from Employee where [IN and OUT Time] in (select [IN and OUT Time] from Employee where Comments= 'Store Open') group by [IN and OUT Time]union all select 'No. Of Emp At Close', count([IN and OUT Time]) from Employee where [IN and OUT Time] in (select [IN and OUT Time] from Employee where Comments= 'Store Close') group by [IN and OUT Time]
答案 2 :(得分：0)
您可以使用此功能：
SELECT
T.IN_OUT_TYPE AS Header1
,(SELECT COUNT(*) FROM tableABC ABC WHERE ABC.IN_and_OUT_ Time = T.IN_and_OUT_ Time) AS Header2
FROM
(
SELECT
'No. Of Emp At Open' AS IN_OUT_TYPE
IN_and_OUT_ Time
FROM tableABC
WHERE Comments = 'Store Open'
UNION ALL
SELECT
'No. Of Emp At Close' AS IN_OUT_TYPE
IN_and_OUT_ Time
FROM tableABC
WHERE Comments = 'Store Close'
) AS T
答案 3 :(得分：0)
试试这个
select [IN and OUT Time],count(*) from table
where [IN and OUT Time] in(select [IN and OUT Time] from table
where Comments in('Store Open','Store Close'))
答案 4 :(得分：0)
没有正确解释。 什么是正确的银泰，包括恩典？
declare @CorrectInTimeIncludingGrace datetime='8:15'
declare @CorrectOutTimeIncludingGrace datetime='22:00'
declare @AttendanceDate datetime='2017-04-24'
declare @AttendanceInDate datetime
declare @AttendanceOutDate datetime
set @AttendanceInDate=dateadd(minute,datepart(minute,@CorrectInTimeIncludingGrace)
, dateadd(hour,datepart(hour,@CorrectInTimeIncludingGrace), @AttendanceDate))
set @AttendanceOutDate=dateadd(minute,datepart(minute,@CorrectOutTimeIncludingGrace)
, dateadd(hour,datepart(hour,@CorrectOutTimeIncludingGrace), @AttendanceDate))
declare @t table(EmployeeID int,IN_OUT dateTime, Type varchar(20),InOutID int, Comments varchar(40))
insert into @t VALUES
(12961,'2017-04-24 08:07:00.000', 'Punch In' , 1 ,'Store Open')
,(12680,'2017-04-24 08:07:00.000','Punch In', 2 , null)
,(12662,'2017-04-24 08:07:00.000','Punch In', 3 , null)
,(12683,'2017-04-24 08:27:00.000','Punch In', 4 , null)
,(12864,'2017-04-24 08:42:00.000','Punch In', 5 , null)
,(12681,'2017-04-24 10:03:00.000','Punch In', 6 , null)
,(-1 ,'2017-04-24 13:33:00.000','Punch In', 7 , null)
,(12662,'2017-04-24 18:00:00.000','Punch Out', 8 , null)
,(12683,'2017-04-24 18:00:00.000','Punch Out', 9 , null)
,(12864,'2017-04-24 18:35:00.000','Punch Out', 10, null)
,(12681,'2017-04-24 22:00:00.000','Punch Out', 11, null)
,(12960,'2017-04-24 22:00:00.000','Punch Out', 12, null)
,(12959,'2017-04-24 22:00:00.000','Punch Out', 13, null)
,(-1 ,'2017-04-24 22:00:00.000','Punch Out', 14, 'Store Close')
select 'No. Of Emp At Open' as Header, count(*)HeaderCount
from @t where
IN_OUT<= @AttendanceInDate
union ALL
select 'No. Of Emp At Close' , count(*)
from @t where
IN_OUT>= @AttendanceOutDate