Access: DateDiff Function

In Access, the DateDiff function returns the number of time intervals between two dates.

The syntax for the DateDiff function is:

DateDiff ( interval, date1, date2[, firstdayofweek[, firstweekofyear]] )

You can use the DateDiff function to determine how many time intervals exist between two dates. For example, you can use DateDiff to calculate the number of days between an order date and its shipdate or the number of weeks between today and the end of the year.  

interval is the interval of time to use to calculate the difference between date1 and date2. The following is a list of valid  interval values.

Interval	Description
yyyy	Year
q	Quarter
m	Month
y	Day of year
w	Weekday
ww	Week
h	Hour
n	Minute
s	Second

date1, date 2 are the two dates to calculate you want to use in the calculation.

firstdayofweek is optional argument that specifies the first day of the week. If not specified, Sunday is assumed. If not specified, the first week is assumed to be the week in which Jan 1 occurs. The following is a list of valid parameters for firstweekofyear.

Constant	Value	Description
vbUseSystem	0	Use the NLS API settings
vbSunday	1	Sunday (default used)
vbMonday	2	Monday
vbTuesday	3	Tuesday
vbWednesday	4	Wednesday
vbThursday	5	Thursday
vbFriday	6	Friday
vbSaturday	7	Saturday

firstweekofyear is optional argument that specifies the first week of the year. The following is a list of valid parameters for firstweekday.

Constant	Value	Description
vbUseSystem	0	Use the NLS API settings
vbFirstJan1	1	Start with the first week in which January 1 occurs (default).
vbFirstFourDays	2	Start with the first week in the year that has at least 4 days
vbFirstFullWeek	3	Start  with the first full week of the year

Examples

DateDiff ('m', #12/17/2003#, #01/22/2004# )           returns 1 DateDiff ('yyyy', #11/05/2001#, #11/10/2003# )        returns 2

VBA Code

Dim MyNumber As Integer  MyNumber = DateDiff('d',#10/06/1998#,#12/09/1998#)

This example uses the DateDiff function to display the number of days between a given date and today. Now the MyNumber variable would contain the value of 64.

SQL query

You can also use the DateDiff function in a query.

SELECT DateDiff('d',#05/05/1985#, Now()) AS Expr1 FROM Orders