Solved

Time Interval

Posted on 2002-04-04
10
256 Views
Last Modified: 2010-05-02
I am getting problem to compare two time with format(hh:mm:ss). I want to know what keyword or function can compare and give the difference between two time interval.
Like comparing 10.12.22 and 10.23.50. The keyword or function should give me the difference between the two times.

buy guy and goodluck.


0
Comment
Question by:mahenky
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
10 Comments
 
LVL 52

Accepted Solution

by:
Ryan Chong earned 50 total points
ID: 6920154
Hi mahenky,

Use the DateDiff function in VB, example:

debug.print datediff("s","10.12.22","10.23.50")
0
 
LVL 6

Expert Comment

by:sharmon
ID: 6920161
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
0
 
LVL 6

Expert Comment

by:sharmon
ID: 6920166
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...
0
Online Training Solution

Drastically shorten your training time with WalkMe's advanced online training solution that Guides your trainees to action. Forget about retraining and skyrocket knowledge retention rates.

 
LVL 43

Expert Comment

by:TimCottee
ID: 6920271
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")
0
 
LVL 6

Expert Comment

by:sharmon
ID: 6921212
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
0
 
LVL 4

Expert Comment

by:RichW
ID: 6921559
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)

0
 
LVL 6

Expert Comment

by:sharmon
ID: 6922061
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
0
 
LVL 4

Expert Comment

by:RichW
ID: 6922241
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

0
 
LVL 6

Expert Comment

by:sharmon
ID: 6922387
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
0
 
LVL 49

Expert Comment

by:DanRollins
ID: 7642937
Hi mahenky,
It appears that you have forgotten this question. I will ask Community Support to close it unless you finalize it within 7 days. I will ask a Community Support Moderator to:

    Accept ryancys's comment(s) as an answer.

mahenky, if you think your question was not answered at all or if you need help, just post a new comment here; Community Support will help you.  DO NOT accept this comment as an answer.

EXPERTS: If you disagree with that recommendation, please post an explanatory comment.
==========
DanRollins -- EE database cleanup volunteer
0

Featured Post

Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Introduction While answering a recent question (http://www.experts-exchange.com/Q_27402310.html) in the VB classic zone, I wrote some VB code in the (Office) VBA environment, rather than fire up my older PC.  I didn't post completely correct code o…
If you need to start windows update installation remotely or as a scheduled task you will find this very helpful.
Get people started with the process of using Access VBA to control Outlook using automation, Microsoft Access can control other applications. An example is the ability to programmatically talk to Microsoft Outlook. Using automation, an Access applic…
Get people started with the utilization of class modules. Class modules can be a powerful tool in Microsoft Access. They allow you to create self-contained objects that encapsulate functionality. They can easily hide the complexity of a process from…

734 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question