How to compare two dates to find time difference in SQL Server 2005, date manipulation

I have two columns:

job_start                         job_end
2011-11-02 12:20:37.247           2011-11-02 13:35:14.613

How would it be possible using T-SQL to find the raw amount of time that has passed between when the job started and when the job ended?

I tried this:

select    (job_end - job_start) from tableA

but ended up with this:

1900-01-01 01:14:37.367

Take a look at the DateDiff() function.

-- Syntax
-- DATEDIFF ( datepart , startdate , enddate )

-- Example usage
SELECT DATEDIFF(DAY, GETDATE(), GETDATE() + 1) AS DayDiff
SELECT DATEDIFF(MINUTE, GETDATE(), GETDATE() + 1) AS MinuteDiff
SELECT DATEDIFF(SECOND, GETDATE(), GETDATE() + 1) AS SecondDiff
SELECT DATEDIFF(WEEK, GETDATE(), GETDATE() + 1) AS WeekDiff
SELECT DATEDIFF(HOUR, GETDATE(), GETDATE() + 1) AS HourDiff
...

You can see it in action / play with it here

You can use the DATEDIFF function to get the difference in minutes, seconds, days etc.

SELECT DATEDIFF(MINUTE,job_start,job_end)

MINUTE obviously returns the difference in minutes, you can also use DAY, HOUR, SECOND, YEAR (see the books online link for the full list).

If you want to get fancy you can show this differently for example 75 minutes could be displayed like this: 01:15:00:0

Here is the code to do that for both SQL Server 2005 and 2008

-- SQL Server 2005
SELECT CONVERT(VARCHAR(10),DATEADD(MINUTE,DATEDIFF(MINUTE,job_start,job_end),'2011-01-01 00:00:00.000'),114)

-- SQL Server 2008
SELECT CAST(DATEADD(MINUTE,DATEDIFF(MINUTE,job_start,job_end),'2011-01-01 00:00:00.000') AS TIME)

Cast the result as TIME and the result will be in time format for duration of the interval.

select CAST(job_end - job_start) AS TIME(0)) from tableA