获取员工工作时间

Question

I have a table with a DATETIME column named CHECKTIME that represents when the employee clocked in/out. Basically, I have a query that does that function but the conversion is not correct. Example below.

CHECKTIME

2014-04-28 07:58:36.000
2014-04-28 12:00:27.000
2014-04-28 13:00:02.000
2014-04-28 16:57:45.000

---

My query result gives me Hour = 7.99 when the real result is Hours = 7.59.

I want the 7.59 result.

SELECT  USERID
       , convert(varchar(10),CHECKTIME,101) as Date
       , SUM(CASE WHEN CHECKTYPE='I' 
                      THEN -1*(datepart(hh,[CHECKTIME])+datepart(mi,[CHECKTIME])/60.0)
                      ELSE datepart(hh,[CHECKTIME])+datepart(mi,[CHECKTIME])/60.0  END) AS Hours

FROM [CHECKINOUT]
WHERE USERID=1003
GROUP BY USERID, CONVERT(varchar(10),[CHECKTIME],101)

Answer 1

Expressing your query like this:

with data(InTime,OutTIme) as (
    select InTime,OutTIme from (values
         (convert(datetime,'2014-04-28 07:58:36.000',120), convert(datetime,'2014-04-28 12:00:27.000',120))
        ,(convert(datetime,'2014-04-28 13:00:02.000',120), convert(datetime,'2014-04-28 16:57:45.000',120))
    )data(InTime,OutTIme)
)
select  cast(dateadd(n,sum(datediff(n,intime,outtime)),0) as time)
from data
go

yields this:

07:59:00.0000000

where I have taken the liberty of pre-pivoting the data manually in order to focus attention on how to construct the time conversion formula.

Update : - amended to include the pivot, generating the same value:

with data(CheckTime) as (
    select CheckTime from (values
         (convert(datetime,'2014-04-28 07:58:36.000',120))
        ,(convert(datetime,'2014-04-28 12:00:27.000',120))
        ,(convert(datetime,'2014-04-28 13:00:02.000',120))
        ,(convert(datetime,'2014-04-28 16:57:45.000',120))
    )data(CheckTime)
),
pvt as (
    select
        CheckInOut
       ,[0]   as OutTime
       ,[1]   as InTime  
    from (
        select 
             CheckTime
            ,(ROW_NUMBER() over(order by CheckTime)-1) / 2 as CheckInOut
            ,(ROW_NUMBER() over(order by CheckTime)) % 2 as rn
        from data
    )t
    pivot (max(CheckTime) for rn in ([1], [0])) pvt

)
select  cast(dateadd(n,sum(datediff(n,intime,outtime)),0) as time)
from pvt
go

Answer 2

You provided a sample query and sample data. But the sample data are incomplete. Please add the values for CHECKTYPE since this will make it a lot easier for us to provide a solution. At the same time it avoids any unnecessary work/guessing like Peter had to do. Finally, a reply like "it doesn't work" is equal to no reply since it doesn't bring us a single step further....

Answer 3

First, lets make some test data: (I took the liberty of adding a emp id column so I could add multiple employee times.

DECLARE @CheckDate TABLE
(
    EmpId INT,
    CheckTime DATETIME
)

INSERT INTO @CheckDate
( EmpId, CheckTime ) 
VALUES
(1, '2014-04-28 07:58:36.000'),
(1, '2014-04-28 12:00:27.000'),
(1, '2014-04-28 13:00:02.000'),
(1, '2014-04-28 16:57:45.000'),
(2, '2014-04-27 1:00:02.000'),
(2, '2014-04-27 16:57:45.000');

Now we make a CTE to find the next row so we can get a begin and end of the clockin and clockout.

;WITH CheckRows AS
(
    SELECT *,
        ROW_NUMBER() OVER (PARTITION BY EmpId ORDER BY CheckTime) RN
    FROM @CheckDate p
),
CheckBreaks AS
(
    SELECT 
        p.EmpId, 
        p.CheckTime, 
        p.RN, 
        ISNULL(n.CheckTime, 0) NextTime
    FROM CheckRows p
    INNER JOIN CheckRows n
        ON n.RN = p.RN + 1
        AND n.EmpId = p.EmpId
)

Now we query the data to find the minutes worked based on the clockins (RN 1 as first clockin/clockout, RN 3 as second clockin/clockout, if you allow for more, just add the next couple of odd numbers for each group)

SELECT 
    c.EmpId,
    SUM(CASE WHEN c.RN = 1 OR c.RN = 3 THEN DATEDIFF(MINUTE, c.CheckTime, c.NextTime) ELSE 0 END) DiffTime,
    SUM(CASE WHEN c.RN = 1 OR c.RN = 3 THEN DATEDIFF(MINUTE, c.CheckTime, c.NextTime) ELSE 0 END) / 60 DiffHour,
    SUM(CASE WHEN c.RN = 1 OR c.RN = 3 THEN DATEDIFF(MINUTE, c.CheckTime, c.NextTime) ELSE 0 END) % 60 DiffMinute
FROM CheckBreaks c
GROUP BY c.EmpId

Here is the output for the hour / minutes worked:

EmpId   DiffTime    DiffHour    DiffMinute
1   479 7   59
2   957 15  57