Solved

# DateDiff Fucntion

Posted on 2004-10-16
612 Views
Hi,

The following code retrive number of days passed between two date.

dateDiff("d",rst("date1"),rst("date2"))

How can I retrieve; suppose the days passed between two date retrive 400 days then I want to display like as following format:

1 year 1 month 5 days

Could anyone help me to do it?

ayha1999
0
Question by:ayha1999
• 5
• 4
• 2
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LVL 95

Expert Comment

ID: 12328749
I'd probably do something like this:

If datediff("y",rst("date1"), rst("date2")) > 0 Then
If datediff("y",rst("date1"), rst("date2")) > 1 Then
sOutput = CStr(datediff("y",rst("date1"), rst("date2"))) & " years "
Else
sOutput = CStr(datediff("y",rst("date1"), rst("date2"))) & " year "
End IF
End IF
If datediff("m",rst("date1"), rst("date2")) > 0 Then
sOutput = sOutput & CStr(datediff("m",rst("date1"), rst("date2"))) & " months "
Else
sOutput = CStr(datediff("m",rst("date1"), rst("date2"))) & " month "
End IF
End If
If datediff("d",rst("date1"), rst("date2")) > 0 Then
sOutput = sOutput & CStr(datediff("d",rst("date1"), rst("date2"))) & " days"
Else
sOutput = CStr(datediff("d",rst("date1"), rst("date2"))) & " day"
End IF
End If

0

LVL 95

Expert Comment

ID: 12328765
Oops, I made a mistake - the "Else ..." sOutput lines for the days and months wouldn't work right - here's the corrected version:

If datediff("y",rst("date1"), rst("date2")) > 0 Then
If datediff("y",rst("date1"), rst("date2")) > 1 Then
sOutput = CStr(datediff("y",rst("date1"), rst("date2"))) & " years "
Else
sOutput = CStr(datediff("y",rst("date1"), rst("date2"))) & " year "
End IF
End IF
If datediff("m",rst("date1"), rst("date2")) > 0 Then
sOutput = sOutput & CStr(datediff("m",rst("date1"), rst("date2"))) & " months "
Else
sOutput = sOutput & CStr(datediff("m",rst("date1"), rst("date2"))) & " month "
End IF
End If
If datediff("d",rst("date1"), rst("date2")) > 0 Then
sOutput = sOutput & CStr(datediff("d",rst("date1"), rst("date2"))) & " days"
Else
sOutput = sOutput & CStr(datediff("d",rst("date1"), rst("date2"))) & " day"
End IF
End If

Incidentally, the resulting string would be in the format you requested.  Note: you could probably turn this into a function to reuse it easier later.
0

LVL 7

Author Comment

ID: 12328886
I tried but I get the following error;

Error Type:
Microsoft VBScript compilation (0x800A0400)
Expected statement
End If

could u pls. check the synatax. some 'ifs' are there without 'end ifs'

ayha
0

LVL 46

Accepted Solution

fritz_the_blank earned 150 total points
ID: 12329264
I wrote some code to show that difference between a given date and Now()--with a little modification, it might work for you:

function timeDifference(datDate)
intDays = DateDiff("d",Now(), datDate)
if intDays >0 then
if Hour(datDate) > Hour(Now()) then
else
intDays = intDays -1
end if
end if
intHours = DateDiff("h",Now(),datDate)
if intHours >0 then
if Minute(datDate) > Minute(Now()) then
else
intHours = intHours -1
end if
end if
intMinutes = DateDiff("n",Now(),datDate)

if intMinutes >0 then
if Second(datDate) > Second(Now()) then
else
intMinutes = intMinutes -1
end if
end if
intSeconds = DateDiff("s",Now(),datDate)

timeDifference = " Days: " & intDays &  " Hours: " & intHours & " Minutes: " &  intMinutes  & " Seconds: " & intSeconds
end function
0

LVL 95

Expert Comment

ID: 12331253
If datediff("y",rst("date1"), rst("date2")) > 0 Then
If datediff("y",rst("date1"), rst("date2")) > 1 Then
sOutput = CStr(datediff("y",rst("date1"), rst("date2"))) & " years "
Else
sOutput = CStr(datediff("y",rst("date1"), rst("date2"))) & " year "
End IF
End IF
If datediff("m",rst("date1"), rst("date2")) > 0 Then
if datediff("m",rst("date1"), rst("date2")) = 1 Then
sOutput = sOutput & CStr(datediff("m",rst("date1"), rst("date2"))) & " month "
Else
sOutput = sOutput & CStr(datediff("m",rst("date1"), rst("date2"))) & " months "
End IF
End If
If datediff("d",rst("date1"), rst("date2")) > 0 Then
If datediff("d", rst("date1"), rst("date2")) = 1 Then
sOutput = sOutput & CStr(datediff("d",rst("date1"), rst("date2"))) & " day"
Else
sOutput = sOutput & CStr(datediff("d",rst("date1"), rst("date2"))) & " days"
End IF
End If

Sorry about that - try the above.  (My server's down at the moment and I can't test things out, but I think I straightened out the bugs).
0

LVL 7

Author Comment

ID: 12379039
@ leew,

I tried your latest code but it gives wrong result;

e.g. I have two date as follows;

5/9/2004  - 5/29/2004 and the result is;

20 years 20 days

could u pls check what's wrong.

@ FTB,

I have not checked ur post so far,I will chekc it later and get back to u.

ayha
0

LVL 95

Assisted Solution

Lee W, MVP earned 100 total points
ID: 12382550
Sorry about that - debugged it in Visual Basic - SHOULD work with ASP:

StartDate = rst("date1")
EndDate = rst("date2")
If DateDiff("yyyy", StartDate, EndDate) > 0 Then
If DateDiff("yyyy", StartDate, EndDate) > 1 Then
sOutput = CStr(DateDiff("yyyy", StartDate, EndDate)) & " years "
Else
sOutput = CStr(DateDiff("yyyy", StartDate, EndDate)) & " year "
End If
End If

NewDate = DateAdd("yyyy", DateDiff("yyyy", StartDate, EndDate), StartDate)
If DateDiff("m", NewDate, EndDate) > 0 Then
If DateDiff("m", NewDate, EndDate) = 1 Then
sOutput = sOutput & CStr(DateDiff("m", NewDate, EndDate)) & " month "
Else
sOutput = sOutput & CStr(DateDiff("m", NewDate, EndDate)) & " months "
End If
End If
NewDate = DateAdd("m", DateDiff("m", StartDate, EndDate), StartDate)
If DateDiff("d", NewDate, EndDate) > 0 Then
If DateDiff("d", NewDate, EndDate) = 1 Then
sOutput = sOutput & CStr(DateDiff("d", NewDate, EndDate)) & " day"
Else
sOutput = sOutput & CStr(DateDiff("d", NewDate, EndDate)) & " days"
End If
End If
0

LVL 7

Author Comment

ID: 12388276
@ leew,

I tried ur latest  but see the result;

I have the following dates;

5/9/2004    5/29/2004

the result is
4 years 5 months 3 days20 days

pls check it?

ayha
0

LVL 95

Expert Comment

ID: 12388918
I should have my server systems running by the end of the weekend.  I'll check it then and let you know.
0

LVL 15

Expert Comment

ID: 12407793
> 4 years 5 months 3 days20 days

The trailing 20 days part seems correct... and it looks tacked onto the end of another value... so just to check... make sure you're not calling leew's code multiple times in the same page and concatenating the results...
0

LVL 7

Author Comment

ID: 12457222
thanks for all
0

LVL 46

Expert Comment

ID: 12457325

FtB
0

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