Sql查询中的时间日志透视

我一直在查询我的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

所需输出：

   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

类型

（1）=超时，而

类型

（2）=超时

员工需要在12点注销，然后在几分钟后重新登录

我试着使用我在这里读到的基于pivot的问题。虽然这是我第一次尝试在表中使用pivot

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;                        

这可能吗？谢谢D

编辑：

正如vercelli所建议的，另一个场景，例如，用户忘记他/她在几分钟前已经计时，所以她再次计时。 e、 g

新查询的输出：

 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    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

Emp RecordDate Time\u in 1 Time\u out 1 Time\u in 2 Time\u out 2
----  ----------    ---------   ---------    --------     ---------
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空我想这就是你要找的

输出

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
6   2016-07-19  07:24:15    12:15:26    12:45:45    17:10:22

我想这就是你要找的

输出

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
6   2016-07-19  07:24:15    12:15:26    12:45:45    17:10:22


---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)


质疑


---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)


质疑

我得到timeout>timein1，应该是另一个way@vercelli：更新支票now@TheGameiswar：感谢您的回答，我已尝试过该查询，但它输出[Err]42000-[SQL Server]“lead”不是可识别的内置函数名。可能是由于不兼容问题。我在微软网站上读到这个函数是从2012年及以上版本开始内置的。我得到的是timeout>timein1，应该是另一个way@vercelli：更新支票now@TheGameiswar：感谢您的回答，我已尝试过该查询，但它输出[Err]42000-[SQL Server]“lead”不是可识别的内置函数名。可能是由于不兼容问题。我在微软网站上读到这个功能是从2012年及以上版本开始内置的。谢谢你的回答。查询是否根据其包含的值过滤结果或行：我试着在这里创建一个例子。例如，用户登录了两次，因为他/她忘记了她刚刚登录。@jkw我明白你的意思，但你会如何显示它？请编辑您的问题以显示此新场景谢谢您的回答。查询是否根据其包含的值过滤结果或行：我试着在这里创建一个例子。例如，用户登录了两次，因为他/她忘记了她刚刚登录。@jkw我明白你的意思，但你会如何显示它？请编辑您的问题以显示此新方案
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;  

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

 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)

 ;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