VBScript » Functions » DateDiff
Version: 2.0
- Syntax:
- DateDiff(Interval, Date1, Date2, FirstDayofWeek, FirstWeekofYear)
- Interval
- The Interval argument defines the the type of time interval you
wish to use to calculate the time difference.
Only the settings from the Interval Settings table may be used, and you must place the setting inside a pair of double quotes.
- Date1
- The Date1 argument is the first date.
- Date2
- The Date2 argument is the second date.
- FirstDayofWeek
- The FirstDayofWeek argument must only use the constants
or values defined below in the Date And Time CONSTANTS.
- FirstWeekofYear
- The FirstWeekofYear argument must only use the constants or values defined in the Date And Time CONSTANTS which are listed above.
The DateDiff function calculates the amount of time between two
different dates.
Interval Settings
| SETTING | DESCRIPTION |
|---|---|
| YYYY | Year |
| Q | Quarter |
| M | Month |
| Y | Day Of Year |
| D | Day |
| W | WeekDay |
| WW | Week Of Year |
| H | Hour |
| N | Minute |
| S | Second |
Date and Time Constants
| CONSTANT | VALUE | DESCRIPTION |
|---|---|---|
| VBSunday | 1 | Sunday |
| VBMonday | 2 | Monday |
| VBTuesday | 3 | Tuesday |
| VBWednesday | 4 | Wednesday |
| VBThursday | 5 | Thursday |
| VBFriday | 6 | Friday |
| VBSaturday | 7 | Saturday |
| VBFirstJan1 | 1 | Week of January 1 |
| VBFirstFourDays | 2 | First week of the year that has at least four days |
| VBFirstFullWeek | 3 | First full week of the year |
| VBUseSystem | 0 | Use the date format of the computer's regionsl settings |
| VBUseSystemDayOfWeek | 0 | Use the first full day of the week as defined by the system settings |
Examples
Code:
<%=DateDiff("d", Date, "1/1/2000") %>Output:
291Language(s): VBScript
Code:
<% olddate = #6/26/43# %>
<% nowdate = Now %>
<% =DateDiff("s", nowdate, olddate) %>Output:
-3130748108Explanation:
The order of the two dates determines if the difference is represented
as a positive or negative value.
Language(s): VBScript
Code:
<% =DateDiff("w", Date, "1/1/2001", 3)
%>Output:
93Explanation:
In this example, Tuesday is defined to be the first day of the week.
The output is how many weeks, which are now defined to start on
Tuesday, are left between the current date and 1/1/2001.
Language(s): VBScript
Code:
<% =DateDiff("w", Date, "1/1/2001", VBTUESDAY)
%>Output:
93Language(s): VBScript
Code:
<% =DateDiff("ww", Date, "1/1/2001", 1,
VBFIRSTFOURDAYS)%>Output:
94Language(s): VBScript
See Also:
