DateDiff
Returns a Variant (Long) specifying the number of time intervals between two specified dates.
Syntax: DateDiff ( interval, date1, date2 [, firstdayofweek [, firstweekofyear ]] )
- interval
- required String expression that is the interval of time used to calculate the difference between date1 and date2. See Interval settings.
- date1, date2
- required Variant (Date). Two dates to use in the calculation.
- firstdayofweek
- optional A VbDayOfWeek constant specifying the first day of the week. Defaults to vbSunday.
- firstweekofyear
- optional A VbFirstWeekOfYear constant specifying the first week of the year. Defaults to vbFirstJan1.
Interval 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 |
firstdayofweek settings
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | 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. |
firstweekofyear settings
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | NLS API setting. |
| vbFirstJan1 | 1 | Week in which January 1 occurs (default). |
| vbFirstFourDays | 2 | First week that has at least four days in the new year. |
| vbFirstFullWeek | 3 | First full week 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, DateDiff 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 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. If date1 or date2 is enclosed in double quotation marks and you omit the year, the current year is inserted each time the expression is evaluated.
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.
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
Dim Msg As String
TheDate = InputBox("Enter a date")
Msg = "Days from today: " & DateDiff("d", Now, TheDate)
MsgBox Msg