Re: Subtracting Dates

"Alan Mills" <Alan.Mills_at_xservices.pants.fujitsu.com> wrote in message news:<bgamaq$1hbb$1_at_news.icl.se>...
> "Ed prochak" <ed.prochak_at_magicinterface.com> wrote in message
> news:4b5394b2.0307301246.2504f1d8_at_posting.google.com...
> > Mark.Powell_at_eds.com (Mark D Powell) wrote in message
> news:<2687bb95.0307281823.506f85f6_at_posting.google.com>...
[]
> > >
> > > Carlos, it is fairly simple. Subtract the older date value from the
> > > newer one. The result is days and fractions there of. Divide the
> > > base (or days) by 24 to get hours. Multiple the fraction by the
> > > number of minutes in a day to get minutes and parts there of.
> > > Multiple the fraction of a minute by 60 to get seconds.
> > >
> > > I am not at work so I do not have a computer available to provide a
> > > partial example, but if you need one after you play around with the
> > > above and no one else posts anything I will post one.
> > >
> > > HTH -- Mark D Powell --
> >
> > Nice little misdirection there Mark. That should make any student
> > programmers scratch their heads. I love it! Regular programmers will
> > see it and say "oh, yeah he meant..."
> >
> > Ed Prochak
>
> Yeah, a bit procedural. IF the difference is less than a day then use
> something like

Carlos must have solved it by now, so I'll explain.

 Actually I was referring to Mark's comment:
> > > .. The result is days and fractions there of. Divide the
> > > base (or days) by 24 to get hours.

It obviously should be MULTIPLY by 24.

 Ed Received on Thu Jul 31 2003 - 13:50:06 CDT