DateDiff
DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]])
Description
DateDiff returns the number of intervals between two dates. The required parameter, interval, is a string expression, see Table 6.. The required parameters, date1 and date2, can be any expressions that represent dates. The optional parameter firstdayofweek is Sunday, if not specified. The optional firstweekofyear is the week containing January 1, if not specified.
Table 7: firstdayofweek constants
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | NLS API setting |
| vbSunday | 1 | Sunday |
| vbMonday | 2 | Monday |
| vbTuesday | 3 | Tuesday |
| vbWednesday | 4 | Wednesday |
| vbThursday | 5 | Thursday |
| vbFriday | 6 | Friday |
| vbSaturday | 7 | Saturday |
Table 8: firstweekofyear constants
| Constant | Value | Description |
|---|---|---|
| vbUseSystem | 0 | NLS API setting |
| vbFirstJan1 | 1 | Week with January 1 |
| vbFirstFourDays | 2 | First week of year with at least 4 days |
| vbFirstFulWeek | 3 | First full week of year |
Example
Rem DateDiff Example
'DateDiff calc difference between 2 dates
Dim Born
Born = InputBox("Enter your birthdate")
Born = CDate(Born)
Print "Since " & Born & " there have been"
Print DateDiff("d", Born, Now) & " days"
Print "or"
Print DateDiff("n", Born, Now) & " minutes"
Output
Since 12/27/1970 there have been
10096 days
or
14539612 minutes
(''sample date output is system dependant'')