DateDiff function
Returns a Variant (Long) specifying the number of time intervals between two specified dates.
Syntax
DateDiff(interval, date1, date2, [ firstdayofweek, [ firstweekofyear ]] )
The DateDiff function syntax has these named arguments:
| Part | Description |
|---|---|
| interval | Required. String expression that is the interval of time you use to calculate the difference between date1 and date2. |
| date1, date2 | Required; Variant (Date). Two dates you want to use in the calculation. |
| firstdayofweek | Optional. A constant that specifies the first day of the week. If not specified, Sunday is assumed. |
| firstweekofyear | Optional. A 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. |
Settings
The interval argument has these settings:
| Setting | Description |
|---|---|
| yyyy | Year |
| q | Quarter |
| m | Month |
| y | Day of year |
| d | Day |
| w | Weekday |
| ww | Week |
| h | Hour |
| n | Minute |
| s | Second |
The firstdayofweek argument has these settings:
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | Use the 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 has these settings:
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | Use the NLS API setting. |
| vbFirstJan1 | 1 | Start with week in which January 1 occurs (default). |
| vbFirstFourDays | 2 | Start with the first week that has at least four days in the new year. |
| vbFirstFullWeek | 3 | Start with first full week of the year. |
Remarks
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.
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 weeks between the two dates. If date1 falls on a Monday, DateDiff counts the number of Mondays until date2. It counts date2 but not date1.
If interval is Week ("ww"), however, the DateDiff function returns the number of calendar weeks between the two dates. It counts the number of Sundays between date1 and date2. DateDiff counts date2 if it falls on a Sunday; but it doesn't count date1, even if it does fall on a Sunday.
If date1 refers to a later point in time than date2, the DateDiff function returns a negative number. 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 double 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.
Note
For date1 and date2, if the Calendar property setting is Gregorian, the supplied date must be Gregorian. If the calendar is Hijri, the supplied date must be Hijri.
Example
This example uses the DateDiff function to display the number of days between a given date and today.
Dim TheDate As Date ' Declare variables.
Dim Msg
TheDate = InputBox("Enter a date")
Msg = "Days from today: " & DateDiff("d", Now, TheDate)
MsgBox Msg
See also
Support and feedback
Have questions or feedback about Office VBA or this documentation? Please see Office VBA support and feedback for guidance about the ways you can receive support and provide feedback.