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 Posted Wednesday, February 2, 2011 9:28 AM
 SSC Rookie Group: General Forum Members Last Login: Thursday, September 24, 2015 1:31 PM Points: 42, Visits: 110
 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
Post #1057562
 Posted Wednesday, February 2, 2011 9:39 AM
 SSC Eights! Group: General Forum Members Last Login: 2 days ago @ 2:19 PM Points: 948, Visits: 7,148
 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`
Post #1057565
 Posted Wednesday, February 2, 2011 10:07 AM
 SSC Rookie Group: General Forum Members Last Login: Thursday, September 24, 2015 1:31 PM Points: 42, Visits: 110
 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.
Post #1057585
 Posted Wednesday, February 2, 2011 10:56 AM
 SSC Eights! Group: General Forum Members Last Login: 2 days ago @ 2:19 PM Points: 948, Visits: 7,148
 Why bother with functions?:`DECLARE @starttime int = 80010 ,@endtime int = 161500SELECT ((@endtime/10000 * 60) + (@endtime%10000/100)) - ((@starttime/10000 * 60) + (@starttime%10000/100))`
Post #1057612
 Posted Wednesday, February 2, 2011 11:04 AM
 SSC Rookie Group: General Forum Members Last Login: Thursday, September 24, 2015 1:31 PM Points: 42, Visits: 110
 Thanks
Post #1057618

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