Time Interval

Here is a routine that I wrote to do pretty much the same thing, you can adjust it to fit your needs.

Public Function DateDiffFractional(ByVal dtStart As Date, ByVal dtEnd As Date) As String
  Dim lngSecs As Long
  Dim lngYears As Long
  Dim lngDays As Long
  Dim lngHours As Long
  Dim lngMinutes As Long
  Dim strTemp As String
 
  If Not IsDate(dtStart) Or Not IsDate(dtEnd) Then
    'Enter your error handling here
    'Invalid date(s)
    Exit Function
  ElseIf CDate(dtEnd) < CDate(dtStart) Then
    'Enter your error handling here
    'Date end is less than date start
    Exit Function
  End If
 
  lngSecs = 0
  lngYears = 0
  lngDays = 0
  lngHours = 0
  lngMinutes = 0
 
  lngSecs = DateDiff("s", dtStart, dtEnd)
   
  If lngSecs >= 31536000 Then
    lngYears = Fix(lngSecs / 31536000)
    lngSecs = lngSecs - (lngYears * 31536000)
  End If
 
  If lngSecs >= 86400 Then
    lngDays = Fix(lngSecs / 86400)
    lngSecs = lngSecs - (lngDays * 86400)
  End If
 
  If lngSecs >= 3600 Then
    lngHours = Fix(lngSecs / 3600)
    lngSecs = lngSecs - (lngHours * 3600)
  End If
 
  If lngSecs >= 60 Then
    lngMinutes = Fix(lngSecs / 60)
    lngSecs = lngSecs - (lngMinutes * 60)
  End If
 
  If lngYears > 0 Then strTemp = "Years: " & lngYears & " "
  If lngDays > 0 Then strTemp = strTemp & "Days: " & lngDays & " "
  If lngHours > 0 Then strTemp = strTemp & "Hours: " & lngHours & " "
  If lngMinutes > 0 Then strTemp = strTemp & "Minutes: " & lngMinutes & " "
  If lngSecs > 0 Then strTemp = strTemp & "Seconds: " & lngSecs
 
  DateDiffFractional = Trim(strTemp)
End Function

BTW: DateDiff will work fine as a single line function if you only need the seconds, else you will have to convert the hours and minutes yourself from the seconds...

sharmon, very true but here is a neat trick to get this done automatically:

Debug.Print format(dateadd("s",datediff("s","10:12:22","10:23:50") ,format(now(),"yyyy-mm-dd")),"ttttt")

Tim,

That's very cool, still left a little formatting, but a cool trick.

As well, that function I gave above can also be shortened down to this for just times...

Public Function TimeDiff(ByVal Time1 As Date, ByVal Time2 As Date) As String
    Dim intHours As Integer
    Dim intMinutes As Integer
    Dim intSeconds As Integer
   
    intSeconds = Abs(DateDiff("s", TimeValue(Time1), TimeValue(Time2)))
   
    If intSeconds >= (60 * 60) Then
        intHours = Fix(intSeconds / (60 * 60))
        intSeconds = intSeconds - (intHours * (60 * 60))
        TimeDiff = Format$(intHours, "00") & ":"
    End If
       
    If intSeconds >= 60 Then
        intMinutes = Fix(intSeconds / 60)
        intSeconds = intSeconds - (intMinutes * 60)
        TimeDiff = TimeDiff & Format$(intMinutes, "00") & ":"
    End If
   
    TimeDiff = TimeDiff & Format$(intSeconds, "00")
End Function

Here's a simpler one:

Public Function TimeDiff(ByVal Interval As String, ByVal Time1 As String, ByVal Time2 As String)
   
    Dim HourTime1, HourTime2, MinuteTime1, MinuteTime2, SecondTime1, SecondTime2, TimeDifference
    Dim Time1InSeconds, Time2InSeconds
    Time1 = FormatDateTime(Time1, vbLongTime)
    Time2 = FormatDateTime(Time2, vbLongTime)
    HourTime1 = Hour(Time1)
    HourTime2 = Hour(Time2)
    MinuteTime1 = Minute(Time1)
    MinuteTime2 = Minute(Time2)
    SecondTime1 = Second(Time1)
    SecondTime2 = Second(Time2)
    Time1InSeconds = SecondTime1 + (MinuteTime1 * 60) + (HourTime1 * 3600)
    Time2InSeconds = SecondTime2 + (MinuteTime2 * 60) + (HourTime2 * 3600)
    TimeDifference = Time2InSeconds - Time1InSeconds
    Select Case Interval
        Case "h"
            TimeDifference = TimeDifference / 3600
        Case "m"
            TimeDifference = TimeDifference / 60
    End Select
    TimeDiff = TimeDifference
End Function

Call it by:
timeNow = Now
timeDifferenceHours = TimeDiff("h", timeStart, timeNow)
timeDifferenceMinutes = TimeDiff("m", timeStart, timeNow)
timeDifferenceSeconds = TimeDiff("s", timeStart, timeNow)

RichW,

How is that simpler and what's the major difference between using what you have there and just the built in DateDiff function?

Debug.Print DateDiff("s", "10:12:22","10:23:50")

I suppose your function would give you fractions of an Interval if needed.

Regards,
Shannon

I thought he wanted just time.

My apologies, Sharmon.  I saw 43 lines of code and missed your last posting. The overhead in If statements and DateDiff's is made up by my variant declarations.  lol

Must be all the beer I had last night.


Regards,
RichW

It's no problem, just thought maybe I was missing something there in your code.

Watch all that beer, you need to save some for the rest of us.

Shannon

Hi mahenky,
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