> -----Original Message----- > From: John Taylor-Johnston [mailto:John.Taylor- > johns...@cegepsherbrooke.qc.ca] > Sent: Saturday, October 16, 2010 10:58 PM > To: PHP-General > Subject: [PHP] Re: strtotime > > According to this, I'm 44 not 45 :)p > > $birthday = '1965-08-30'; > > //calculate years of age (input string: YYYY-MM-DD) > > function birthday ($birthday){ > list($year,$month,$day) = explode("-",$birthday); > $year_diff = date("Y") - $year; > $month_diff = date("m") - $month; > $day_diff = date("d") - $day; > if ($day_diff < 0 || $month_diff < 0)
I think the problem is with the logic above. $year_diff = 44 $month_diff = 2 $day_diff = -13 Since the $month_diff is positive meaning you're already passed your birthdate. If the $month_diff is 0 - your birth month, then you need to check if the if it's past your birth day yet (negative $day_diff). That logic should be changed to if if (($month_diff == 0 && $day_diff < 0) || $month_diff < 0) Then you're 45 ;) Regards, Tommy > $year_diff--; > return $year_diff; > } > > echo birthday ($birthday); > > > > John Taylor-Johnston wrote: > > I'm working my way right now through the manual: > > http://ca.php.net/manual/en/function.strtotime.php. > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
