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SSC Rookie
      
Group: General Forum Members
Last Login: Friday, May 10, 2013 6:09 AM
Points: 42,
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Hi,
I have the following integer 80010, All I want to do is convert this to 08:00:10 so I can use in a DateDiff function.
Is there a time function which will do this or do I have to resort to breaking the number down and appending the ':'?
Thanks
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SSC Eights!
      
Group: General Forum Members
Last Login: Yesterday @ 5:19 AM
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Maybe:
DECLARE @time int, @dt datetime; SET @time = 80010; SET @dt = DATEADD(second, 80010%100,DATEADD(minute, 80010%10000/100,DATEADD(hour, 80010/10000,'19000101')));
SELECT @dt
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SSC Rookie
      
Group: General Forum Members
Last Login: Friday, May 10, 2013 6:09 AM
Points: 42,
Visits: 99
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Thanks. I already have a date in a different column and I trying to calculate the number of minutes worked by an employee on a given date. I have the starttime as 80010 (integer) and endtime as 161500. The result I am looking for is 495 minutes. I believe I can achieve this result if I was able to convert startdate 80010 to 08:00:10 and endate to 16:15:00 and then using the datediff function to get the results in minutes.
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SSC Eights!
      
Group: General Forum Members
Last Login: Yesterday @ 5:19 AM
Points: 822,
Visits: 5,103
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Why bother with functions?:
DECLARE @starttime int = 80010 ,@endtime int = 161500
SELECT ((@endtime/10000 * 60) + (@endtime%10000/100)) - ((@starttime/10000 * 60) + (@starttime%10000/100))
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SSC Rookie
      
Group: General Forum Members
Last Login: Friday, May 10, 2013 6:09 AM
Points: 42,
Visits: 99
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Thanks
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