Time Calculations 1

[cell0042]

I want to be able to do a time calculation that will not let the minutes go over 60. Right I have the get the difference of two times to be something like 10.75 hrs. I want it to be like 10 and 45 minutes. Can I please get some help!!!!!

 

[allanon]

I wrote you a little function to do what you want. Pass it a double and it will return a string. Modify it any way you need to.

Public Function ConvertToTime(InNumber As Double) As String
Dim hours As Integer
Dim minutes As Integer

hours = CInt(Mid(Format(InNumber, "###0.00&quot, 1, Len(Format(InNumber, "###0.00&quot) - 3))

minutes = (InNumber * 100) Mod 100

minutes = minutes * 60 / 100

ConvertToTime = CStr(hours) & " Hours " & CStr(minutes) & " Minutes"
End Function


call it like so:

Me.txtTimeOut = ConvertToTime(CDbl(Me.txtTimeIn))

 

[grnzbra]

Don't know if there is an "official" way to do it but, you could create a function that takes in the two times.

Within the function:
1. Determine the difference in minutes. (DateDiff())

2. Divide this by 60.

3. Move the result of Step 2 (eg.10.75) to a text variable.

3. Take the Left(Variable,Len(variable)-3) of this text variable. (Should be 10)

4. Take the Mod of the division of the difference in minutes by 60. The Mod function gives the "remainder" of a division. Since the division is by 60, it gives the number of 60ths of an hour, ie minutes.

5. Concatinate the two with a colin between them and set the function equal to that.

 

[sdk]

Hi

Some options a little easier (& with ref to MichaelRed):

If the period is always less than one day, use:
= format(LaterTime - EarlierTime,"hh:mm&quot

If the period can be mulitple days, use
= DateDiff("d",EarlierTime,LaterTime) & " Days, " & format(LaterTime - EarlierTime,"hh:mm&quot

These can be used directly in the properties of the contol on your form or report.

Cheers

Steve