我在查询我的timelogs表时有点困惑， 以下是样本数据：
TimelogId EmployeeId RecordDate RecordTime Type
--------- ---------- ---------- ---------- ----
1 4 2016-07-01 07:18:37 1
2 4 2016-07-01 12:03:14 2
5 4 2016-07-01 12:08:02 1
6 4 2016-07-01 18:02:03 2
7 6 2016-07-19 07:24:15 1
8 5 2016-07-19 07:26:03 1
9 6 2016-07-19 12:15:26 2
10 5 2016-07-19 12:35:17 2
13 5 2016-07-19 12:36:14 1
16 6 2016-07-19 12:45:45 1
17 6 2016-07-19 17:10:22 2
18 5 2016-07-19 18:43:09 2
所需的输出：
Date EmployeeId Time_In Time_Out Time_In Time_Out
------- ---------- -------- -------- ------- -------
2016-07-01 4 07:18:37 12:03:14 12:08:03 18:02:03
2016-07-19 6 07:24:15 12:15:26 12:45:45 17:10:22
2016-07-19 5 07:26:03 12:35:17 12:36:14 18:43:08
其中Type（1）=时间和Type（2）=超时。
员工必须在12nn注销，然后在几分钟后重新登录。
我尝试使用基于我在此阅读的先前问题的数据透视表。虽然这是我第一次尝试在表格中使用数据透视。
select
*
FROM
(
select
EmployeeID,
RecordDate,
RecordTime,
Type
from tblTimeLogs
) d
PIVOT(
Max(RecordTime) <---- I think the problem is right around here
FOR Type in ([1],[2]) <----- plus i can't seem to rename this column names maybe i'll read a little further on this.
) piv;
查询输出：
EmployeeID RecordDate 1 2
---------- ---------- ------ ------
4 2016-07-01 12:08:02 18:02:03
5 2016-07-19 12:36:14 18:43:09
6 2016-07-19 12:45:45 17:10:22
修改
正如vercelli所建议的那样，另一种情况就是例如用户忘了他/她已经在几分钟前定时，所以她再次进入。 e.g。TimelogId EmployeeId RecordDate RecordTime Type
--------- ---------- ---------- ---------- ----
1 4 2016-07-01 07:18:37 1
2 4 2016-07-01 12:03:14 2
5 4 2016-07-01 12:08:02 1
6 4 2016-07-01 18:02:03 2
7 6 2016-07-19 07:24:15 1
8 5 2016-07-19 07:26:03 1
9 6 2016-07-19 12:15:26 2
10 5 2016-07-19 12:35:17 2
13 5 2016-07-19 12:36:14 1
16 6 2016-07-19 12:45:45 1
17 6 2016-07-19 17:10:22 2
18 5 2016-07-19 18:43:09 2
19 5 2016-07-20 08:13:35 1 <--- Time in
20 5 2016-07-20 08:14:35 1 <--- Timed In again
21 5 2016-07-20 12:15:12 2 <--- Time Out
我尝试使用Vercelli先生的询问：
select
EmployeeID as Emp, RecordDate, [1] as Time_In1, [2] as Time_Out1, [3] as Time_In2, [4] as Time_out2
FROM
(
select
EmployeeID,
RecordDate,
RecordTime,
ROW_NUMBER () over (partition by EmployeeId, RecordDate order by RecordTime, type) as rn
from tblTimeLogs
) d
PIVOT(
max(RecordTime)
FOR rn in ([1],[2],[3],[4])
) as piv;
输出新查询：
Emp RecordDate Time_In1 Time_Out1 Time_In2 Time_Out2
---- ---------- --------- --------- -------- ---------
4 2016-07-01 07:18:37 12:03:14 12:08:02 18:02:03
5 2016-07-19 07:26:03 12:35:17 12:36:14 18:43:09
5 2016-07-20 08:13:35 08:14:35 12:15:12 Null <--- the problem is right around this portion
6 2016-07-19 07:24:15 12:15:26 12:45:45 17:10:22
预期产出：
Emp RecordDate Time_In1 Time_Out1 Time_In2 Time_Out2
---- ---------- --------- --------- -------- ---------
4 2016-07-01 07:18:37 12:03:14 12:08:02 18:02:03
5 2016-07-19 07:26:03 12:35:17 12:36:14 18:43:09
5 2016-07-20 08:13:35 12:15:12 08:14:35 Null <--- the second time in would fall on the second 5th column (Time_In2) since the **Type** value is = 1
6 2016-07-19 07:24:15 12:15:26 12:45:45 17:10:22
感谢帮助人员：D。我正在从这个问题中学习新事物。
答案 0 :(得分：0)
我认为这就是你要找的东西。 Demo：
select
EmployeeID as Emp, RecordDate, [1] as Time_In1, [2] as Time_Out1, [3] as Time_In2, [4] as Time_out2
FROM
(
select
EmployeeID,
RecordDate,
RecordTime,
ROW_NUMBER () over (partition by EmployeeId, RecordDate order by RecordTime, type) as rn
from tblTimeLogs
) d
PIVOT(
max(RecordTime)
FOR rn in ([1],[2],[3],[4])
) as piv;
<强> OUPUT
Emp RecordDate Time_In1 Time_Out1 Time_In2 Time_out2
4 2016-07-01 07:18:37 12:03:14 12:08:02 18:02:03
5 2016-07-19 07:26:03 12:35:17 12:36:14 18:43:09
6 2016-07-19 07:24:15 12:15:26 12:45:45 17:10:22
答案 1 :(得分：0)
---表创建脚本
create table #test
(
employeeid int,
recorddate date,
recordtime time,
typee int
)
insert into #test values( 4 , '2016-07-01', '07:18:37', 1)
insert into #test values( 4 , '2016-07-01', '12:03:14', 2)
insert into #test values( 4 , '2016-07-01', '12:08:02', 1)
insert into #test values( 4 , '2016-07-01', '18:02:03', 2)
查询
;with cte
as
(
select
employeeid,
max(recorddate) as recorddate,
min(recordtime) as timein,
max(recordtime) as Timein1 ,
lead(min(recordtime)) over (partition by employeeid order by min(recordtime)) as 'timeout1',
lead(max(recordtime)) over (partition by employeeid order by max(recordtime)) as 'timeout2'
from #test
group by
employeeid,typee
)
select employeeid,recorddate,timein,timeout1,timein1,timeout2
from cte
where timeout1 is not null and timeout2 is not null