2017-03-08 50 views
1

我有一种情况,员工有时间和超时,但他们被保存到一个类型为[in = 1,out = 2],我需要使用Time-In,Time-Out标题以单行形式提取单个员工时间(in-out)。如何从SQlLSERVER中的多行子集构建单行

下面是桌子,我有为例,

CREATE TABLE #Employee 
(
empid int, 
name varchar(20), 
age int 
) 
CREATE TABLE #TIMEINOUT 
(
    id int, 
    empid int, 
    timeinout datetime, 
    [type] tinyint 
) 

INSERT INTO #Employee values(1,'Benny',35) 
INSERT INTO #Employee values(2,'Algo',32) 


INSERT INTO #TIMEINOUT VALUES(1,1,'2017-03-08 06:00:00 AM',1) -- (Type 1 = IN , 2 = Out) 
INSERT INTO #TIMEINOUT VALUES(2,1,'2017-03-08 05:00:00 PM',2) -- (Type 1 = IN , 2 = Out) 
INSERT INTO #TIMEINOUT VALUES(3,2,'2017-03-08 07:00:00 AM',1) -- (Type 1 = IN , 2 = Out) 
INSERT INTO #TIMEINOUT VALUES(4,2,'2017-03-08 09:00:00 PM',2) -- (Type 1 = IN , 2 = Out) 



SELECT * FROM #Employee INNER JOIN #TIMEINOUT ON #Employee.empid = #TIMEINOUT.empid 

Select #Employee.empid,#Employee.name,#Employee.age,GETDATE() as TimeIN,GETDATE() as TimeOUT from #Employee 

DROP TABLE #Employee 
DROP TABLE #TIMEINOUT 

任何人可以帮助简化查询?

+0

预期的输出应该是 – Chanukya

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@ Chanukya..I已编辑并选择了预期结果 – DareDevil

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您需要通过empid和日期来做到这一点? –

回答

0

这是我如何做,

SELECT #Employee.empid , 
    #Employee.name , 
    #TIMEINOUT.timeinout [TimeIN], 
    X.timeinout [TimeOUT] 
FROM #Employee 
    INNER JOIN #TIMEINOUT ON #TIMEINOUT.empid = #Employee.empid 
    LEFT JOIN (SELECT #TIMEINOUT.empid , 
         #TIMEINOUT.timeinout 
       FROM #TIMEINOUT 
       WHERE type = 2 
      ) X ON X.empid = #Employee.empid 
        AND CONVERT(VARCHAR(10), #TIMEINOUT.timeinout, 111) = CONVERT(VARCHAR(10), X.timeinout, 111) 
WHERE #TIMEINOUT.type = 1 
1

也许这样?为TimeinTimeOut

;with cte as (
SELECT e.empid,e.name,e.age,t.timeinout as TimeIn FROM #Employee e INNER JOIN #TIMEINOUT t ON e.empid = t.empid 
where t.type=1 
) 
select t.empid,t.name,t.age,t.timein as TimeIn,t2.timeinout as Timeout from cte t 

inner join (
select * from #timeinout 
where type=2 
)t2 
on t.empid=t2.empid 
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没有结果t2无效 – DareDevil

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他刚刚错过了t2.empid – Pream

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这是否适用于同一用户的多次进出时间? – Pream

1

单独列看看这个:

select a.empid,a.timeinout,b.timeinout from 
(select *,CONVERT(VARCHAR(10),timeinout,110) as this_in from TIMEINOUT where type = 1) as a 
inner join 
(select *,CONVERT(VARCHAR(10),timeinout,110) as this_out from TIMEINOUT where type = 2) as b 
on a.empid = b.empid where a.this_in = b.this_out 
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如果您想要员工姓名,只需加入员工。 –

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伟大的方法,我只是想选择所有,然后选择一个列与左连接具有相同的emp_id和相同的日期与差异是类型为超时 – DareDevil

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日期是重要的,这就是为什么我用它包括在where子句中只是为了使肯定 –