How to compute the duration of time

• • Reply

  14 years ago

  by niko66623

  niko ilustre

  , Guam

  Joined 14 years ago

  hello please help me anyone the problem is i know how to copute for the duartion of 2 time like for example a login and a log out and there duration but i dont know how to compute if the log out is more than 1 day and week

  please give me the code for computing duration for more than 1 day or 1 week here is the sample code

  Dim Time1 As Date

  Dim Time2 As Date

  Dim Dur As Date

  get the time first to be stored in text1 and text2

  Time1 = CDate(Text1.Text)

  Time2 = CDate(Text2.Text)

  Dur = Time2 - Time1

  Text3.Text = Format(Dur,"hh:mm:ss")

  this code computes duration for just 1 day and i want to know how to compute for more than a day or a week please help me please......

  Report abuse
• • Reply

  14 years ago

  by khari6579

  Hari K

  Chennai, India

  Joined 15 years ago

  Hi Niko
   I herewith post coding for duration calculation.
   I hope this will helpful for u
  
   u just open one VB-Form add three TextBox text1,text2,text3 and add one more command button name as command1.
  
  Finally copy and paste the follwoing code
  Run the form and click the command button no need to give input.
  This is for Example i am posting here
  
  Private Sub Command1_Click()
  Dim Dt1, Dt2 As Date
  Dim strHour As String
  Dim strMin As String
  Dim strSec As String
  
  
  Text1.Text = "09-Jun-2006 05:00:10"
  Text2.Text = "13-Jun-2006 23:56:59"
  Dt1 = CDate(Text1.Text)
  Dt2 = CDate(Text2.Text)
  strSec = DateDiff("s", Dt1, Dt2)
  
  If CLng(strSec) > 0 Then
      strHour = Int(CLng(strSec) / 3600)
      strSec = CLng(strSec) Mod 3600
      If CLng(strSec) > 0 Then
          strMin = Int(CLng(strSec) / 60)
          strSec = CLng(strSec) Mod 60
      End If
  End If
  
  If Len(strHour) = 1 Then
   strHour = "0" & strHour
  End If
  
  If Len(strMin) = 1 Then
   strMin = "0" & strMin
  End If
  
  If Len(strSec) = 1 Then
   strSec = "0" & strSec
  End If
  
  Text3.Text = strHour & ":" & strMin & ":" & strSec
  End Sub
  
  Regards
  Hari K
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Report abuse
• • Reply

  14 years ago

  by sync_or_swim

  Rob Bickel

  S. Wales, United Kingdom

  Joined 15 years ago

  In the example given, the output would be: 114:56:49.  There are two ways in which you can extend Hari's code to break down the output into days and hours.
  
  Get answer as whole number of hours and divide by 24 hours in a day:
  This is an example of how you can calculate the number of days in a given number of hours by dividing the number of hours by 24, just take out the line that reads Text3.Text = strHour & ":" & strMin & ":" & strSec and cut and paste the following in its place:
  
  Dim counter As Integer
  Dim whole As Integer
  Dim remainder As Integer
  whole = 0
  If strHour > 24 Then
      For counter = strHour - 24 To 0 Step -24
          whole = whole + 1
          If counter < 24 Then
              remainder = counter
              counter = counter - remainder
          End If
      Next
  Else
     remainder = strHour
  End If
  If whole > 0 Then
      Text3.Text = whole & "Days " & remainder & "Hours " & strMin & "Mins " & strSec & "Secs"
  Else
      Text3.Text = remainder & "Hours " & strMin & "Mins " & strSec & "Secs"
  End If
  
  Calculate number of whole days between the dates, then add this number to date1:
  This is a bit slicker, just calculate the difference between the two dates using "d" as the interval, this will give you a whole number of days.  Once you know the number of days you can add this to date1 and perform the datediff calculation again to find the number of hours:
  
  Private Sub Command1_Click()
  Dim Dt1, Dt2 As Date
  Dim strHour As String
  Dim strMin As String
  Dim strSec As String
  Dim nofdays As String
  Text1.Text = "09-Jun-2006 05:00:10"
  Text2.Text = "13-Jun-2006 23:56:59"
  Dt1 = CDate(Text1.Text)
  Dt2 = CDate(Text2.Text)
  
  'Calculation 1
  noofdays = DateDiff("d", Dt1, Dt2)
  
  'Take number of days off
  Dt1 = DateAdd("d", noofdays, Dt1)
  
  'Now do second calculation
  strSec = DateDiff("s", Dt1, Dt2)
  If CLng(strSec) > 0 Then
      strHour = Int(CLng(strSec) / 3600)
      strSec = CLng(strSec) Mod 3600
      If CLng(strSec) > 0 Then
          strMin = Int(CLng(strSec) / 60)
          strSec = CLng(strSec) Mod 60
      End If
  End If
  If Len(strHour) = 1 Then
   strHour = "0" & strHour
  End If
  If Len(strMin) = 1 Then
   strMin = "0" & strMin
  End If
  If Len(strSec) = 1 Then
   strSec = "0" & strSec
  End If
  
  Text3.Text = noofdays & " " & strHour & ":" & strMin & ":" & strSec
  End Sub
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Report abuse
• • Reply

  14 years ago

  by niko66623

  niko ilustre

  , Guam

  Joined 14 years ago

  thanks to all of you guys you saved my life the 2 of you i will never forget the two of you

  thank you very much.....

  Report abuse
• • Reply

  14 years ago

  by ErwinB

  Erwin Bergsma

  Leuven, Belgium

  Joined 15 years ago

  To use less code, you can also use the following:

  iDays = CInt(Val(strHour)/24)
  iHours = Val(strHour) Mod 24
  
  
  

   

  Report abuse