Setting the concatenation on the display is good.
The problem that I am having is my startDate is 3/27/2000
3:00:00 PM my endDate is 3/27/2000 3:59:00 PM. The result
that the code is giving me is 0 -1:1, when the result
should be 0 00:59.
If the time difference is longer, say startTime is
2/25/2003 4:00:00 PM and end time is 2/25/2003 7:59:00 PM,
the result is 0 4:-1.
From your expression, I have set dt1 = endDate and dt2 =
startDate. If I reverse these, the result is -1 0:1, so I
think that setting the endDate to dt1 and startDate to dt2
is correct.
Any ideas?
Thanks
-----Original Message-----
Hi,
All you have to do is alter the format within the function.
To get it to return dd hh:mm, just make this change:
ShowDateDiff = lngDays & " " & lngHours & ":" & lngMintues
Similarily, just add more logic to deal with lngDays being 0.
You have the 3 values to work with: lngDays,lngHours and lngMintues.
You can display them any way you please.
--
HTH
Dan Artuso, Access MVP
Thanks, it works, but I would like to have the format as
dd hh:mm. With the dd (number of days) to be null if the
date/time difference is less that 23:59 hours:minutes.
The desired format is the default Short time with the
addition of the number of days in front of the
hours/minutes.
What you have shared with me so far is great.
Thanks - J
-----Original Message-----
Hi,
Not sure if this is what you want but it will return a
string in the format:
12 days 3 hours 0 minutes
Public Function ShowDateDiff(dt1 As Date, dt2 As Date) As
String
Dim lngDays As Long
Dim lngHours As Long
Dim lngMinutes As Long
lngMinutes = DateDiff("n", dt2, dt1)
lngDays = lngMinutes / 1440
lngHours = (lngMinutes - (lngDays * 1440)) / 60
lngMinutes = lngMinutes - (lngDays * 1440) - (lngHours *
60)
ShowDateDiff = lngDays & " days " & lngHours & " hours "
& lngMinutes & " minutes"
End Function
--
HTH
Dan Artuso, Access MVP
I am having difficulty with subtracting two date/times
and
getting the result in the format that will show the
number
of days, the hours and the minutes.
My example is:
(3/13/2003 11:00:00 AM) - (3/1/2003 8:00:00 AM) Which
should result in 12 days 3 hours, but only gives 3 hours
as the difference.
[txtEndTime]-[txtStartTime] Both txtEndTime and
txtStartTime are stored as General Date.
I have tried several methods, including Format
([txtEndDate]-1-[txtStartDate], "Short Time"), which
give
the correct number of hours and minutes, but does not
account for the number of days.
Any help is appreciated
.
.
