MS Access: DateDiff Function
The Microsoft Access DateDiff function returns the difference between two date values, based on the interval specified.
DateDiff Function (Syntax)
The syntax for the DateDiff function is:
DateDiff ( interval, date1, date2, [firstdayofweek], [firstweekofyear])
interval is the interval of time to use to calculate the difference between date1 and date2. Below is a list of valid interval values.
|y||Day of year|
date1 and date2 are the two dates to calculate the difference between.
firstdayofweek is optional. It is a constant that specifies the first day of the week. If this parameter is omitted, Access assumes that Sunday is the first day of the week. This parameter can be one of the following values:
|vbUseSystem||0||Use the NLS API setting|
firstweekofyear is optional. It is a constant that specifies the first week of the year. If this parameter is omitted, Access assumes that the week containing Jan 1st is the first week of the year. This parameter can be one of the following values:
|vbUseSystem||0||Use the NSL API setting|
|vbFirstJan1||1||Use the first week that includes Jan 1st (default)|
|vbFirstFourDays||2||Use the first week in the year that has at least 4 days|
|vbFirstFullWeek||3||Use the first full week of the year|
The DateDiff function can be used in the following versions of Microsoft Access:
- Access 2013, Access 2010, Access 2007, Access 2003, Access XP, Access 2000
|DateDiff ("yyyy", #15/10/1998#, #22/11/2003#)||would return 5|
|DateDiff ("m", #15/10/2003#, #22/11/2003#)||would return 1|
|DateDiff ("d", #15/10/2003#, #22/11/2003#)||would return 38|
DateDiff Examples (in VBA Code)
The DateDiff function can be used in VBA code in Microsoft Access. For example:
Dim LValue As Integer LValue = DateDiff ("d", #15/10/2003#, #22/11/2003#)
In this example, the variable called LValue would now contain the value of 38.
DateDiff Examples (in SQL/Queries)
You can also use the DateDiff function in a query in Microsoft Access.