[AccessD] Subtract Date/Time Fields

Ha!  That's straight into my Goodie Bag Drew.
Thanks
Max





On Wed, Dec 17, 2008 at 3:24 AM, Drew Wutka <DWUTKA at marlow.com> wrote:

> Just subtract them.  Dates are technically double floating point
> numbers.  The whole number is the number of days since Dec 30th 1899 (or
> something like that).  The decimal is the fraction of the day.
>
> So, to determine the number of days, take the result and use Fix to get
> the number of days (as a date, the result would show up as a date, like
> Feb. 22nd).  Then use the Hours, Minutes and Seconds function to get the
> rest.
>
> So, here's an example:
>
> Function GetDateDifference(dtStart As Date, dtEnd As Date) As String
> Dim dtResult As Date
> Dim intTemp As Long
> dtResult = dtEnd - dtStart
> intTemp = Fix(dtResult)
> GetDateDifference = intTemp & " Days " & Hour(dtResult) & " Hours " & _
> Minute(dtResult) & " Minutes " & Second(dtResult) & " Seconds"
> End Function
>
> Drew
> -----Original Message-----
> From: accessd-bounces at databaseadvisors.com
> [mailto:accessd-bounces at databaseadvisors.com] On Behalf Of ACTEBS
> Sent: Tuesday, December 16, 2008 7:38 PM
> To: access group
> Subject: [AccessD] Subtract Date/Time Fields
>
> Hi Everyone,
>
> This is driving me to drink!!
>
> How do you subtract 2 date/time values? eg. 17/12/2008 12:00:00   from
> 20/02/2009 19:40:00
>
> All I am after is how many days, hours, minutes and seconds there are
> between the 2 dates.
>
> Anyone had any experience with this type of problem?
>
> Thanks in advance...
>
> Vlad
>
>
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