DateDiff Function
Syntax
DateDiff(interval, dateexpr1, dateexpr2)
Group
Description
Return the number of intervals between two dates.
|
Parameters |
Description |
|
This string value indicates which kind of interval to subtract. |
|
|
Calculate the interval from this date value to dateexpr2. If this value is Null then Null is returned. |
|
|
Calculate the interval from dateexpr1 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
Example
Sub Main
Debug.Print DateDiff("yyyy",#1/1/1990#,#1/1/2000#) ' 10
End Sub