Subtracting Time 1

[jass1220]

what shall i do if i want to find out the time duration ...


txtFrom = 9:00

txtTo = 3:00

lblDuration = txtTo - txtFrom



thanks

 

[CCLINT]

For minutes use:

?(CDate("9:30&quot-CDate("9:00&quot)*1440

For Hours use:

?(CDate("9:30&quot-CDate("9:00&quot)* 24

I would use the above method especially in a loop where many recodes are being checked.

Otherwise, you can use DateDiff():

?DateDiff("n", "9:00", "9:30&quot

 

[Genimuse]

txtFrom = "9:00"
txtTo = "3:00"

lblDuration = DateDiff("h", CDate(txtFrom), CDate(txtTo)) & _
 ":" & DateDiff("n", CDate(txtFrom), CDate(txtTo))

Should to the trick.

 

[Schweiger]

Try this:

lblDuration = cDate(cDate(txtTo) - cDate(txtFrom))

or to keep formatting:

lblDuration = format((cDate(txtTo) - cDate(txtFrom)),"h:mm")

or have a look at DateDiff

hope this helps
Andreas

 

[CCLINT]

Genimuse, that will not work...

 

[jass1220]

txtFrom = "9:00"
txtTo = "3:00"

lblduration= CDate(CDate(txtTo) - CDate(txtFrom))

but the result was in this formate 6:00:00 AM


i want it to be 6


i want to calculate the number of hours ...

 

[johnwm]

Did you try CCLINT's solution?


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[CCLINT]

I did mention this:

?(CDate("9:30&quot-CDate("9:00&quot)*1440

But you need to use Date Variables for the time or use 24 hour format, as 9:00 is strictly 9:00 AM.

Same as doing this:

CDate("3:00&quot which is always 3:00 AM.

The same will happen with
lblduration= CDate(CDate(txtTo) - CDate(txtFrom))

You can see this clearly on a machine set in 24 hour time format.

So, everything should work fine if you either use a 24 hour time format when working with strings, or use a date variable.

This is where it is a must to store time in a db field in a date/time field, and not in a text field, unless you have a 24 hour identifier such as 2 digit hours or using the AM/PM identifier.
Same holds true for user input.
Who knows what they mean when they input 3:00, unless they can select AM/PM or know that 3:00 is really 03:00.

jass1220, you need to be a little more observant.

Schweiger did give you a second choice so you only needed to use that, or a variant of that, such as using just "h".
For this, the Hour() or Minute function would be better though.

 

[CCLINT]

Thanks johnwm.

jass1220, let me say again: You will run into problems using:

txtFrom = "9:00"
txtTo = "3:00"

While this is 6 hours either way, changing txtTo = "4:00" is going to give you 5 hours and not the "expected" 7 hours!

For that you need to do this:
txtTo = "16:00"
or
txtTo = "4:00 PM"

 

[jass1220]

oppps

thanks programers .. working