DateDiff Function
Returns the number of intervals between two dates.
Syntax
DateDiff(interval, date1, date2 [,firstdayofweek[, firstweekofyear]])
Arguments
interval
Required. String expression that is the interval you want to use to calculate the differences between date1 and date2. See Settings section for values.date1, date2
Required. Date expressions. Two dates you want to use in the calculation.firstdayofweek
Optional. Constant that specifies the day of the week. If not specified, Sunday is assumed. See Settings section for values.firstweekofyear
Optional. Constant that specifies the first week of the year. If not specified, the first week is assumed to be the week in which January 1 occurs. See Settings section for values.
Settings
The interval argument can have the following values:
Setting |
Description |
|---|---|
yyyy |
Year |
q |
Quarter |
m |
Month |
y |
Day of year (same as Day) |
d |
Day |
w |
Weekday |
ww |
Week of year |
h |
Hour |
n |
Minute |
s |
Second |
The firstdayofweek argument can have the following values:
Constant |
Value |
Description |
|---|---|---|
vbUseSystemDayOfWeek |
0 |
Use National Language Support (NLS) API setting. |
vbSunday |
1 |
Sunday (default) |
vbMonday |
2 |
Monday |
vbTuesday |
3 |
Tuesday |
vbWednesday |
4 |
Wednesday |
vbThursday |
5 |
Thursday |
vbFriday |
6 |
Friday |
vbSaturday |
7 |
Saturday |
The firstweekofyear argument can have the following values:
Constant |
Value |
Description |
|---|---|---|
vbUseSystem |
0 |
Use National Language Support (NLS) API setting. |
vbFirstJan1 |
1 |
Start with the week in which January 1 occurs (default). |
vbFirstFourDays |
2 |
Start with the week that has at least four days in the new year. |
vbFirstFullWeek |
3 |
Start with the first full week of the new year. |
Remarks
You can use the DateDiff function to determine how many specified time intervals exist between two dates. For example, you might use DateDiff to calculate the number of days between two dates, or the number of weeks between today and the end of the year.
If date1 refers to a later time than date2, the DateDiff function returns a negative number.
To calculate the number of days between date1 and date2, you can use either Day of year ("y") or Day ("d").
When interval is Weekday ("w"), DateDiff returns the number of full 7-day periods between date1 and date2.
When interval is Week ("ww"), however, DateDiff returns the number of Sundays that occur starting one day after date1 through date2. The date1 parameter is not counted if it is a Sunday, but date2 is counted if it is a Sunday.
The firstdayofweek argument affects calculations that use the "w" and "ww" interval symbols.
If date1 or date2 is a date literal, the specified year becomes a permanent part of that date. However, if date1 or date2 is enclosed in quotation marks (" ") and you omit the year, the current year is inserted in your code each time the date1 or date2 expression is evaluated. This makes it possible to write code that can be used in different years.
When comparing December 31 to January 1 of the immediately succeeding year, DateDiff for Year ("yyyy") returns 1 even though only a day has elapsed.
The following example uses the DateDiff function to display the number of days between a given date and today:
Function DiffADate(theDate)
DiffADate = "Days from today: " & DateDiff("d", Now, theDate)
End Function
Requirements
Change History
Date |
History |
Reason |
|---|---|---|
September 2009 |
Modified the remarks. |
Content bug fix. |