I want Hours,Min, second difference from two datetime

I came across an easier way of solving this issue.

First, a quick example of turning a "number of seconds" into the "hh:mm:ss" format.

DECLARE @NumberOfSeconds int
SET @NumberOfSeconds = 3843     --  1 hour, 4 minutes, 3 seconds

SELECT @NumberOfSeconds AS 'seconds',
   CONVERT(varchar, DATEADD(second, @NumberOfSeconds, 0), 108) AS 'hh:mm:ss'

This will give us this output:

enter image description here

And we can easily take this a step further, calculate the number of seconds between two datetimes, and display it in hh:mm:ss format:

DECLARE 
    @NumberOfSeconds int,
    @StartTime datetime = '2017-09-14 14:16:11',
    @EndTime datetime = '2017-09-14 14:23:13'

SET @NumberOfSeconds = DATEDIFF(second, @StartTime, @EndTime)

SELECT @NumberOfSeconds AS 'seconds',
     CONVERT(varchar, DATEADD(second, @NumberOfSeconds, 0), 108) AS 'hh:mm:ss'

Which gives us this output:

enter image description here

Simple, hey ?

(And yes, you can simplify it further by putting the DATEDIFF directly into the DATEADD function.)


You can do it in a very simple way:

declare  @date1 datetime, @date2 datetime
set @date1=DATEADD(s,-638,getdate())
set @date2=GETDATE()

select convert(char(8),dateadd(s,datediff(s,@date1,@date2),'1900-1-1'),8)

... the result is 00:10:38 (638s = 600s + 38s = 10 minutes and 38 seconds)

Another example:

select distinct convert(char(8),dateadd(s,datediff(s, CRDATE , GETDATE() ),'1900-1-1'),8) from sysobjects order by 1

It will works until the difference of 86399 seconds (23:59:59):

select convert(char(8),dateadd(s,datediff(s
    , DATEADD(s,-86399,GETDATE())
    , GETDATE()
),'1900-1-1'),8)

... after that it will return to zero:

select convert(char(8),dateadd(s,datediff(s
    , DATEADD(s,-86400,GETDATE())
    , GETDATE()
),'1900-1-1'),8)