This is what help says for DateDiff

DateDiff Function
Returns a Variant (Long) specifying the number of time intervals between two
specified dates.

Syntax

DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]])

The DateDiff function syntax has these named arguments:

Part Description
interval Required. String expression that is the interval of time you
use to calculate the difference between date1 and date2.
date1, date2 Required; Variant (Date). Two dates you want to use in
the calculation.
firstdayofweek Optional. A constant that specifies the first day of
the week. If not specified, Sunday is assumed.
firstweekofyear Optional. A constant that specifies the first week of
the year. If not specified, the first week is assumed to be the week in
which January 1 occurs.



Settings

The interval argument has these settings:

Setting Description
yyyy Year
q Quarter
m Month
y Day of year
d Day
w Weekday
ww Week
h Hour
n Minute
s Second



The firstdayofweek argument has these settings:

Constant Value Description
vbUseSystem 0 Use the NLS API setting.
vbSunday 1 Sunday (default)
vbMonday 2 Monday
vbTuesday 3 Tuesday
vbWednesday 4 Wednesday
vbThursday 5 Thursday
vbFriday 6 Friday
vbSaturday 7 Saturday



Constant Value Description
vbUseSystem 0 Use the NLS API setting.
vbFirstJan1 1 Start with week in which January 1 occurs (default).
vbFirstFourDays 2 Start with the first week that has at least four
days in the new year.
vbFirstFullWeek 3 Start with first full week of the year.



Remarks

You can use the DateDiff function to determine how many specified time
intervals exist between two dates. For example, you might use DateDiff to
calculate the number of days between two dates, or the number of weeks
between today and the end of the year.

To calculate the number of days between date1 and date2, you can use either
Day of year ("y") or Day ("d"). When interval is Weekday ("w"), DateDiff
returns the number of weeks between the two dates. If date1 falls on a
Monday, DateDiff counts the number of Mondays until date2. It counts date2
but not date1. If interval is Week ("ww"), however, the DateDiff function
returns the number of calendar weeks between the two dates. It counts the
number of Sundays between date1 and date2. DateDiff counts date2 if it falls
on a Sunday; but it doesn't count date1, even if it does fall on a Sunday.

If date1 refers to a later point in time than date2, the DateDiff function
returns a negative number.

The firstdayofweek argument affects calculations that use the "w" and "ww"
interval symbols.

If date1 or date2 is a date literal, the specified year becomes a permanent
part of that date. However, if date1 or date2 is enclosed in double
quotation marks (" "), and you omit the year, the current year is inserted
in your code each time the date1 or date2 expression is evaluated. This
makes it possible to write code that can be used in different years.

When comparing December 31 to January 1 of the immediately succeeding year,
DateDiff for Year ("yyyy") returns 1 even though only a day has elapsed.

Note For date1 and date2, if the Calendar property setting is Gregorian,
the supplied date must be Gregorian. If the calendar is Hijri, the supplied
date must be Hijri.


--

HTH

RP
(remove nothere from the email address if mailing direct)


"Bernard Liengme" wrote in message
...
It is 'hidden' for reasons know only to Mr Gates & Associates. It was
mentioned in the Help file of one version (XL 2000 I believe) but it is
available in all versions. Look at Chip's site www.cpeason.com for details
on using it.
best wishes
--
Bernard V Liengme
www.stfx.ca/people/bliengme
remove caps from email

"Nigel Welch" <Nigel wrote in message
...
I have all Add-ins loaded, but still this function is not in the various
function lists