Note to using the following all you need to is turn your individual date items
into a single string and make the date1 and then using the Now( ) function
as date2.
VBA DateDiff Function
The VB DateDiff function returns a variant variable result of subtracting two
dates. The unit of measure is defined in the interval argument.
Syntax
DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]])
Arguments:
interval - Required. String expression that is the interval of time you use to
calculate the difference between date1 and date2. See a list of valid intervals
in the DateAdd function above.
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.
Example:
MsgBox DateDiff("yyyy", "01/01/2005", "11/11/2006")
Above displays '1'.
Your specific example assumes your items are numeric
strBirthDate = Format(Day_Birth, "00") & "/" & Format(Month_Birth, "00") & "/"
& cStr(Year_Birth)
intAge = DateDiff("yyyy", strBirthDate, Now( ))
--- On Thu, 5/28/09, abdul daulay <[email protected]> wrote:
>
> I want to make a school database, but I don't know how to get a student age.
> I have
> three field name=Day_Birth, Month_Birth and Year_Birth. Please help me...
>
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