Time Difference

[Hillary]

I want to evaluate the difference between two times. The Time In and Time Out formats are hh:mm:ss.

What is the best way to calculate the time difference?

Thanks,

Hillary

 

[jtseltmann]

Check out the DateDiff function.

e.g.
duration = DateDiff("h",starttime,endtime)

 

[Hillary]

I figured out the formula...I converted everything into seconds. How do I convert back to hh:mm:ss?

Thanks!

Hillary

 

[jtseltmann]

There may be a better way but this should work assuming you don't ever have an end time after a start time.

Dim vTimeDuration as variant
Dim vHours As Variant
Dim vMinutes As Variant
Dim vSeconds As Variant

duration = DateDiff("s", starttime,endtime)

vHours = Format(Int(duration \ 3600), "00")
vMinutes = Format((duration Mod 3600) \ 60, "00")
vSeconds = Format((duration Mod 3600) Mod 60, "00")

vTimeDuration = vHours & ":" & vMinutes & ":" & vSeconds

 

[ESquared]

Keep in mind that DateDiff works with the number of time-interval boundaries crossed between two times. It doesn't actually subtract the two.

That is,

[tt]DateDiff("m", starttime, endttime)[/tt]

will give different results than

[tt]DateDiff("m", 0, endtime-starttime)[/tt]

For the first way, the difference between 12:00:59 and 12:01:00 is 1 (a minute boundary has been crossed). The second way, the difference is 0 (less than a minute has elapsed).

You are using seconds, so the point may be moot here, but if you switch to a database such as SQL Server which stores datetime values with a resolution of 1/300th of a second you'd find your DateDiff function suddenly behaving differently.

Just like the Access example with minutes above, in SQL Server,

[tt]DateDiff(s, '12:00:00.997', '12:01:01.000')[/tt]

will give 1 second difference, even though only 3 milliseconds have passed.

 

[SkipVought]

and...

the UNITS of Time is DAYS, if you're considering converting Seconds to Time (Days).


Skip,

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