DateDiff Function

Syntax:

DateDiff(interval, dateexpr1, dateexpr2)

Group:

Time/Date

Description:

 

Return the number of intervals between two dates.

 

Parameter

Description

interval

This string value indicates which kind of interval to add.

number

Add this many intervals. Use a negative value to get an earlier date.

dateexpr

Calculate the new date relative to this date value. If this value is Null then Null is returned.

 

Interval

Description

yyyy

Year

q  

Quarter  

m  

Month  

y

Day of year

d

Day

w

Weekday

ww

Week

h

Hour

n

Minute

s

Second

 

See Also:

DateAdd, DatePart.

Example:


Sub Main
  Debug.Print DateDiff("yyyy",#1/1/1990#,#1/1/2000#) ' 10
End
Sub