DateDiff Function

Returns the number of date or time intervals between two given date values.

Syntax:


DateDiff (interval As String, date1 As Date, date2 As Date [, firstDayOfWeek As Integer [, firstWeekOfYear As Integer]]) As Double

Return value:

A number.

Parameters:

interval - A string expression from the following table, specifying the date or time interval.

interval (string value)

Explanation

yyyy

Year

q

Quarter

m

Month

y

Day of year

w

Weekday

ww

Week of year

d

Day

h

Hour

n

Minute

s

Second


date1, date2 - The two date values to be compared.

Date literals allow to specify unambiguous date variables that are independent from the current language. Literals are enclosed between hash signs #. Possible formats are:

firstdayofweek: An optional parameter that specifies the starting day of a week.

firstdayofweek value

Explanation

0

Use system default value

1

Sunday (default)

2

Monday

3

Tuesday

4

Wednesday

5

Thursday

6

Friday

7

Saturday


firstweekofyear: An optional parameter that specifies the starting week of a year.

firstweekofyear value

Explanation

0

Use system default value

1

Week 1 is the week that includes 1st January (default)

2

Week 1 is the first week containing four or more days of that year

3

Week 1 is the first week containing only days of the new year


Example:


Sub example_datediff
    MsgBox DateDiff("d", #1/1/2005#, #2005-12-31#)
End Sub

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