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Wildhoney · 09-21-2007, 07:27 PM

I tried that, Village Idiot, but all I appear to get back is 1991:

PHP Code:


			
echo bday('10-10-1985');

function bday($bday)
{
    $age = time() - strtotime($bday);
    $years = date("Y",$age);
    return $years;
}

09-21-2007, 07:27 PM	#3 (permalink)
Wildhoney La Vida es Sueño Join Date: Sep 2007 Location: Oldham Posts: 2,267 Thanks: 90	I tried that, Village Idiot, but all I appear to get back is 1991: PHP Code: `echo bday('10-10-1985'); function bday($bday) { $age = time() - strtotime($bday); $years = date("Y",$age); return $years; }` __________________ The man who comes back through the Door in the Wall will never be quite the same as the man who went out.