Math with Time var's

Hello
I am working on a Time Clock Program and I can find the difference between the clock in time and the clock out time. Is there a vb function to do this? or any thoughts?
My Dream Function would be

HoursWorked = SomeFunction(ClockIn,ClockOut)

If Life could be so good?

Thank you

William

It's named Datediff
DateDiff("h", date1, date2)

Hi William

Here's a silly one. It won't work if the month rolls-over though. Should be fairly simple to fix.

Make a form with a Command button called Command1, then

Code:

---

Dim T1 As Date

Function SomeFunction(ByVal InTime As Date, ByVal OutTime As Date) As String
    'Returns the difference between InTime and OutTime (in minutes)
    InTime = Day(InTime) * 1440 + Hour(InTime) * 60 + Minute(InTime)
    OutTime = Day(OutTime) * 1440 + Hour(OutTime) * 60 + Minute(OutTime)
    SomeFunction = OutTime - InTime
End Function

Private Sub Command1_Click()
    ttl = SomeFunction(T1, Now)
    m$ = "Hours: " & vbTab & ttl \ 60 & vbCrLf
    m$ = m$ & "Minutes: " & vbTab & ttl Mod 60
    MsgBox m$, vbInformation, "Time Taken"
End Sub

Private Sub Form_Load()
    T1 = Now
End Sub

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Well, that saves a lot of time. I didn't know about the DateDiff function. Thanks Kedaman

Cool, thanks Andrew
But What is The Mod keyword for?
and I didn't know about the DateDiff function either.

William

The Mod keyword is used to divide two numbers and only return the remainder.

FWIW, another variation on the theme

Code:

---

Function SomeOtherFunction(ClockIn As Date, ClockOut As Date) As Single
Dim Oops As Date
' Return difference between InTime and OutTime (in hours)
  If ClockIn > ClockOut THEN
    Oops = ClockIn
    ClockIn = ClockOut
    ClockOut = Oops
  End If
  SomeOtherFunction = 1440 * (ClockOut - ClockIn)
End Function

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[Edited by Mongo on 07-01-2000 at 05:01 AM]