:12

"Jeff Johnson [MVP: VB]" wrote:

>
> "Paulymon" <Pauly***@discussions.microsoft.com> wrote in message
> news:EC777025-45BC-44CB-9BEE-BD6888DB26BF@microsoft.com...
>
> >I entered the following in my immediate window and expected to see a result
> >of
> > 4:00
> >
> > Why is it showing me :12 ?  I want to subtract the 2 date/times and get 4
> > hours as the result.
>
> Then use DateDiff().
>
> > Please help
> >
> > ?format(#2/2/2006 2:35:00 PM# -#2/2/2006 10:35:00 AM#,"[h]:mm")
> > :12
>
> mm = two-digit month, not minutes. nn = minutes. Why 12, though? Because the
> result of #2/2/2006 2:35:00 PM# - #2/2/2006 10:35:00 AM# is, numerically,
> 0.166666666671517, and since the integer portion of this number (which
> represents the date) is 0, that equates to 1899-12-30, hence the month of
> 12.
>
>
>

"Rick Rothstein [MVP - Visual Basic]" <rickNOSPAMnews@NOSPAMcomcast.net>
wrote in message news:eD4GhwDKGHA.3728@tk2msftngp13.phx.gbl...
>> Thanks for the explanation, how do I get the result (4:00) that I want
>> though.  I tried using nn and get 0
>
> Convert your subtraction to a date so VB knows what it is...
>
> Format(CDate(#2/2/2006 2:35:00 PM# - #2/2/2006 10:35:00 AM#), "hh:nn")
>
> Of course, in a real program, those two dates would be stored in Date
> variables and their subtraction would be stored in a Date variable and
> that
> last Date variable would be used as the first argument.
>
> Rick

> >> Thanks for the explanation, how do I get the result (4:00) that I want
> >> though.  I tried using nn and get 0
> >
> > Convert your subtraction to a date so VB knows what it is...
> >
> > Format(CDate(#2/2/2006 2:35:00 PM# - #2/2/2006 10:35:00 AM#), "hh:nn")
> >
> > Of course, in a real program, those two dates would be stored in Date
> > variables and their subtraction would be stored in a Date variable and
> > that
> > last Date variable would be used as the first argument.
> >
> > Rick
>
> .....and might even use the built in DateDiff function to do the math <g>