在Sql查询中透视时间日志
[英]Pivoting Timelogs in Sql Query
问题描述
I am kind of stuck in querying my timelogs table, the following are the sample data: 
我有点困在查询我的timelogs表，以下是示例数据：
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
The required output: 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
Where Type (1) = time in and Type (2) = time out. 其中Type （1）= time in和Type （2）= time out。
Employees are required to logout at 12nn then log back in again after a couple of minutes. 员工需要在12nn注销，然后在几分钟后重新登录。
I tried using pivot based from the previous questions that i read here. 我尝试使用基于我在此阅读的先前问题的数据透视表。 Though this is the first time i tried to use pivot in tables. 虽然这是我第一次尝试在表格中使用数据透视表。
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;
Output of query: 输出查询：
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
is this possible? 这可能吗？ Thanks. 谢谢。 :D ：d
Edit: 编辑：
as suggested by vercelli, another scenario like for example the user forgot that he/she already timed in just a couple of minutes ago so she timed in again. 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。 eg 例如
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
I tried using the query of Mr. Vercelli: 我尝试使用Vercelli先生的查询：
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;
Output of the new query: 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
Expected Output: 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
Thanks for the help guys :D. 谢谢你的帮助：D。 I'm learning new things from this problem. 我正在从这个问题中学到新东西。
2 个解决方案
解决方案1
0 2016-07-18 09:12:08
I think this is what you are looking for. 我想这就是你要找的东西。 Demo : 演示 ：
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;
OUPUT 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
解决方案2
0 2016-07-18 09:15:32
---Table Creattion Scripts ---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)
Query 询问
;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
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