I know there is a datediff function, you may be able to use it for time also.
I would look into that before using 2 different integers for your time keeping function.
A quick search of the Access help system shows this.
Calculating with dates and times You can add and subtract dates and times and include them in other calculations. To use a date or time in a formula, enter the date or time as text and enclose the text in quotation marks. For example, the following formula would display a difference of 68:
I think if you use Excel you can set the numbers to be valued as hours/minutes.
To get only the remainder back, you're looking for the MOD function.
dblResults = dblValueA MOD dblValueB
dblResults = 3 MOD 2
dblResults = .5
Haven't used VBScript for a while but I believe the above should work.
and =([minutes] mod 60)
Search the MS site...they may have a solution.
You can use the following logic to calculate the difference between two times which works even across midnight.
Format([StartTime] -1 -[EndTime], "Short Time")
A little more info.
If you want to include midnight in your calculation then an easy way is to include a midnight operator (1).
Thus #23.45# -1 = 00:15:00
f you need to go over midnight then you need to include the operator (and format the result to understand it)
#23:45# - 1-#00.15# = -2.08333333333333E-02
format(#23:45#-1-#00:15#,"short time") = 00:30
format(-2.08333333333333E-02, "short time")= 00:30"
Seems you need to use the format "short time" function to change your integer to a date/time format
So your function is format("return time" - 1 -"start time", "short time")
edit: fuckit too tired to get it right