Php date functions

[rshadarack]

I'm trying to use the php date functions, but not for the current year. For example, is there a function I could call to get if 1976 was a leap year? Or to find the first day (Monday, Tuesday, etc) of the year for 1996?

[Houdini]

For checking dates you could use checkdate() [a href=\"http://us3.php.net/manual/en/function.checkdate.php\" target=\"_blank\"]PHP:Manual[/a] and to see the day of the week you could use
[code]$date="1996-01-01";
$mynewdate =date("D j F y",strtotime($date));
echo $mynewdate;[/code]

[neylitalo]

A year has to meet the following requirements to be a leap year:[list][*]It CANNOT be evenly divisible by 100[list][*]Exception: If the year is ALSO evenly divisible by 400, it is a leap year.[/list][*]It MUST be evenly divisible by 4[/list]This function takes care of the requirements.

[code]function isLeapYear($year)
{
    if( (($year % 100 != 0) && ($year % 4 == 0)) || ($year % 400 == 0))
    {
        $isLeapYear = true;
    }
    else
    {
        $isLeapYear = false;
    }
    return $isLeapYear;
}[/code]

[Pastulio]

I don't quite get how it works... I've came to something like that. And now I'm thinking, how did I come up with this because 2000 / 4 = 500 and 2000 / 400 = 5, and none of them is "0".

So I came to this conclusion:

[code]
// Check if the entered year is a leapyear
if (!is_int($_POST['BoughtAtYear'] % 100) && (is_int($_POST['BoughtAtYear'] % 400) || is_int($_POST['BoughtAtYear'] % 4))){

  // In this case the year would be a leapyear

} else {

  // In this case the year would not be a leapyear

}

[/code]

Please tell me where I am wrong here :s

EDIT: Sorry I just saw how stupid I was. the "%" indicator gives the rest and not the value X_X