N.B.: iMonth is zero based, so 0 represents January, 1 represents February, 2 represents March and 11 represents December. iYear is not zero based, this is the actual calendar year number. (2006 is actually 2006)
Pray that the browser developers know the correct way to determine whether a year is a leap year! (It's more complicated than a simple mod 4 == 0) Here is a quote from Wikipedia's page on leap years: "The Gregorian calendar, the current standard calendar in most of the world, adds a 29th day to February in all years evenly divisible by 4, except for centennial years (those ending in '00'), which receive the extra day only if they are evenly divisible by 400. Thus 1600, 2000 and 2400 are leap years but 1700, 1800, 1900 and 2100 are not."
To test this function, February in the years 2100, 2005, 2004, 2003, 2001, 2000 and 1999 should be checked. All of these should return 28, except for 2004 and 2000.
How does this function work? It is quite simple. When the Date() function is given a day number that is greater than the number of days in the given month of the given year, it wraps the date into the next month. The getDate() function returns the day of the month, starting from the beginning of the month that the date is in. So, day 32 of March is considered to be day 1 of April. Subtracting 1 from 32 gives the correct number of days in March!
|All times are GMT +5.5. The time now is 20:45.|