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Old November 20th, 2003, 08:20 AM
ASHY ASHY is offline
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How Do I Use The DateDiff Function To Work Out Someone's Age

Hi Iam very new to Visual Basic 6.0 and have a problem and would be most grateful for some help.

It is as follows:

I have to create a program to input a person's date of birth and output their age in terms of years

I was told I would have to use the DateDiff Function

DateDiff(interval,date1,date2)

I Think Date 1 would be a variable (the users date of birth) and date 2 another variable (the current date), but what is interval is it years and is that what I would type within there.

I know this is probably simple to most of use but again Iam new to programming many thanks in advance

Ashy

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Old November 20th, 2003, 09:35 AM
cleverpig cleverpig is offline
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DateDiff Function
Returns the number of intervals between two dates.

DateDiff(interval, date1, date2 [,firstdayofweek[, firstweekofyear]])
The DateDiff function syntax has these parts:

Arguments
interval
Required. String expression that is the interval you want to use to calculate the differences between date1 and date2. See Settings section for values.
date1, date2
Required. Date expressions. Two dates you want to use in the calculation.
firstdayofweek
Optional. Constant that specifies the day of the week. If not specified, Sunday is assumed. See Settings section for values.
firstweekofyear
Optional. 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. See Settings section for values.
Settings
The interval argument can have the following values:

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

The firstdayofweek argument can have the following values:

Constant Value Description
vbUseSystemDayOfWeek 0 Use National Language Support (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

The firstweekofyear argument can have the following values:

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

The following example uses the DateDiff function to display the number of days between a given date and today:

Function DiffADate(theDate)
DiffADate = "Days from today: " & DateDiff("d", Now, theDate)
End Function
Requirements
Version 2

Plz see msdn page http://msdn.microsoft.com/library/d...fctdatediff.asp

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