MS Access DateDiff() Function
Example
Return the difference between two dates (in years):
SELECT DateDiff("yyyy", #13/01/1998#, #09/05/2017#);
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Definition and Usage
The DateDiff() function returns the difference between two date values, based on the interval specified.
Syntax
DateDiff(interval, date1, date2, firstdayofweek,
firstweekofyear)Parameter Values
| Parameter | Description |
|---|---|
| interval | Required. The interval to use to calculate the difference between date1
and date2. Can be one of the following values:
|
| date1 and date2 | Required. The two dates to calculate the difference between |
| firstdayofweek | Optional. Specifies the first day of the week. If omitted, Sunday is the
first day of the week.
Can be one of the following values:
|
| firstdayofyear | Optional. Specifies the first week of the year. If omitted, the week
containing January 1st is the first week of year. Can be one of the following values:
|
Technical Details
| Works in: | Access 2016, Access 2013, Access 2010, Access 2007, Access 2003, Access XP, Access 2000 |
|---|
More Examples
Example
Return the difference between two dates (in months):
SELECT DateDiff("m", #13/01/1998#, #09/05/2017#);
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Example
Return the difference between a specified date and today's date (in days):
SELECT DateDiff("d", #13/01/1998#, Date());
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