Calculating Time Difference using DateDiff Function in VBScript

[nilesh]

Hello Friends,

I am trying to find difference between two time values. I am using the code
given below.

<!-- #Include file="ADOVBS.INC" -->
<%
xx = "23:55"
yy = "22:00"
xxx = FormatDateTime(xx, vbShortTime)
yyy = FormatDateTime(yy, vbShortTime)
response.write (xxx)
response.write (yyy)
zz=DateDiff(h, xxx, yyy)
response.write (zz)
%>

But I get the error as below:

23:5522:00
Microsoft VBScript runtime error '800a0005'

Invalid procedure call or argument: 'DateDiff'

Can anybody help??

[François Desfossés]

maybe those little code will help ...

Function DateDiffInMinutes(Date1, Date2)

Dim m_yerDate1
Dim m_yerDate2
Dim m_monDate1
Dim m_monDate2
Dim m_dayDate1
Dim m_dayDate2
Dim m_hrDate1
Dim m_hrDate2
Dim m_minDate1
Dim m_minDate2
Dim m_secDate1
Dim m_secDate2
Dim m_minElapsed1
Dim m_monthdays
Dim m_minElapsed2

m_yerDate1 = CInt(Year(Date1)) - 1992 - 1
m_yerDate2 = CInt(Year(Date2)) - 1992 - 1
m_monDate1 = CInt(Month(Date1)) - 1
m_monDate2 = CInt(Month(Date2)) - 1
m_dayDate1 = CInt(Day(Date1)) - 1
m_dayDate2 = CInt(Day(Date2)) - 1
m_hrDate1 = CInt(Hour(Date1))
m_hrDate2 = CInt(Hour(Date2))
m_minDate1 = CInt(Minute(Date1))
m_minDate2 = CInt(Minute(Date2))
m_secDate1 = CInt(Second(Date1))
m_secDate2 = CInt(Second(Date2))
m_minElapsed1 = YearInDays(m_yerDate1) * 1440
m_monthdays = MonthInDays(m_monDate1, IsLeapYear(m_yerDate1 + 1))
m_minElapsed1 = m_minElapsed1 + m_monthdays * 1440
m_minElapsed1 = m_minElapsed1 + m_dayDate1 * 1440
m_minElapsed1 = m_minElapsed1 + m_hrDate1 * 60 + m_minDate1
m_minElapsed2 = YearInDays(m_yerDate2) * 1440
m_monthdays = MonthInDays(m_monDate2, IsLeapYear(m_yerDate2 + 1))
m_minElapsed2 = m_minElapsed2 + m_monthdays * 1440
m_minElapsed2 = m_minElapsed2 + m_dayDate2 * 1440
m_minElapsed2 = m_minElapsed2 + m_hrDate2 * 60 + m_minDate2

g_DateDiff = m_minElapsed2 - m_minElapsed1

' Return 1 for success
DateDiffInMinutes = 1
End Function
----------------------------------------------------------------------------
---------------------------------

--
François Desfossés
Programmeur stagiaires, systèmes bureautiques
Trainee programmer, office systems
Technologie de l'information/Information Technologies
Agence spatiale canadienne/Canadian Space Agency
francois....@space.gc.ca
Http://www.space.gc.ca

[Kevin O'Brien]

Try.......

<%
xx = TimeValue("23:50")
yy = TimeValue("22:00")
response.write (xx & "<BR>")
response.write (yy & "<BR>")
zz=(Hour(xx)-Hour(yy))*60 + (Minute(xx)-Minute(yy))
zz=TimeSerial(0,zz,0)
response.write zz
%>

Depending on your app, you may have to adjust for +/- time values.

nilesh <nilesh.s...@timesgroup.com> wrote in message
news:#XVr3Opz#GA.310@cppssbbsa04...

[Robert Zaufall]

Hi,

you have to put " ' "s around the parameter for the intervall.
Something like datediff('h', date1, date2)

Bye.
Robert

On Thu, 15 Jul 1999 12:59:23 +0530, "nilesh" <nilesh.s...@timesgroup.com> wrote:

>Hello Friends,
>
>...
>zz=DateDiff(h, xxx, yyy)
>...