VBScript DateDiff Function

DateDiff(interval,date1,date2[,firstdayofweek[,firstweekofyear]])

Can take the following values:

• yyyy - Year
• q - Quarter
• m - Month
• y - Day of year
• d - Day
• w - Weekday
• ww - Week of year
• h - Hour
• n - Minute
• s - Second

Can take the following values:

• 0 = vbUseSystemDayOfWeek - Use National Language Support (NLS) API setting
• 1 = vbSunday - Sunday (default)
• 2 = vbMonday - Monday
• 3 = vbTuesday - Tuesday
• 4 = vbWednesday - Wednesday
• 5 = vbThursday - Thursday
• 6 = vbFriday - Friday
• 7 = vbSaturday - Saturday

Can take the following values:

• 0 = vbUseSystem - Use National Language Support (NLS) API setting
• 1 = vbFirstJan1 - Start with the week in which January 1 occurs (default)
• 2 = vbFirstFourDays - Start with the week that has at least four days in the new year
• 3 = vbFirstFullWeek - Start with the first full week of the new year

document.write(Date & "<br />")
document.write(DateDiff("m",Date,"12/31/2002") & "<br />")
document.write(DateDiff("d",Date,"12/31/2002") & "<br />")
document.write(DateDiff("n",Date,"12/31/2002"))

Output:

1/14/2002
11
351
505440

document.write(Date & "<br />")
'Note that in the code below
'is date1>date2
document.write(DateDiff("d","12/31/2002",Date))

Output:

1/14/2002
-351

'How many weeks (start on Monday),
'are left between the current date and 10/10/2002
document.write(Date & "<br />")
document.write(DateDiff("w",Date,"10/10/2002",vbMonday))

Output:

1/14/2002
38