want to get the difference in hh.mm

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want to get the difference in hh.mm

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pnasz
Posting Yak Master

101 Posts

Posted - 2010-12-08 : 08:18:00

want to get the difference in hh.mm can anyone help

(case when (0)>((CONVERT([float],datediff(minute,[EmpSIN],[EmpSOut])/(60),(0))+CONVERT([float],(datediff(minute,[EmpSIN],[EmpSOut])%(60))/(100.0),(0)))-(CONVERT([float],datediff(minute,[FirstIn],[FirstOut])/(60),(0))+CONVERT([float],(datediff(minute,[FirstIn],[FirstOut])%(60))/(100.0),(0)))) then (0) else (CONVERT([float],datediff(minute,[EmpSIN],[EmpSOut])/(60),(0))+CONVERT([float],(datediff(minute,[EmpSIN],[EmpSOut])%(60))/(100.0),(0)))-(CONVERT([float],datediff(minute,[FirstIn],[FirstOut])/(60),(0))+CONVERT([float],(datediff(minute,[FirstIn],[FirstOut])%(60))/(100.0),(0))) end)

jimf
Master Smack Fu Yak Hacker

2875 Posts

Posted - 2010-12-08 : 09:25:19

Not sure what you're after as you didn't post sample data with expected output, but this may help you.

declare @startDate datetime
declare @endDate datetime

set @startDate = current_timestamp

set @endDate = dateadd(minute,abs(checksum(newid()))/1000000,@startdate)

select @startdate
,@enddate
,datediff(minute,@startdate,@enddate)
,datediff(minute,@startdate,@enddate) / 60
,datediff(minute,@startdate,@enddate) % 60
,(datediff(minute,@startdate,@enddate) % 60)/60.0
,datediff(minute,@startdate,@enddate) / 60 + (datediff(minute,@startdate,@enddate) % 60)/60.0

Jim

Everyday I learn something that somebody else already knew

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2010-12-08 : 09:26:43

SELECT CONVERT(CHAR(5), DATEADD(MINUTE, DATEDIFF(MINUTE, EmpSin, EmpSout), 0), 108),
CONVERT(CHAR(5), DATEADD(SECOND, DATEDIFF(SECOND, EmpSin, EmpSout), 0), 108)

N 56°04'39.26"
E 12°55'05.63"

SwePeso
Patron Saint of Lost Yaks

30421 Posts

Posted - 2010-12-11 : 07:48:18

SELECT REPLACE(CONVERT(CHAR(5), DATEADD(SECOND, DATEDIFF(SECOND, EmpSin, EmpSout), 0), 108), ':', '.')

N 56°04'39.26"
E 12°55'05.63"

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