datetime - Age calculation leap year issue in php -

i using following function calculate age given date of birth, not showing correct difference if leap year day i.e 29 used. please me fix code.

<?php function getabsage($birthday)     {         list($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)         {             $year_diff--;         }          if ($year_diff == 0)         {             $interval = date_diff(date_create(), date_create($birthday));             $months = $interval->format("%m");             $days = $interval->format("%d");              if ($months > 0)             {                 return $interval->format("%m months %d days");             }             else if ($months == 0 && $days > 1)             {                 return $interval->format("%d days");             }             else             {                 return $interval->format("%d day");             }         }         else if ($year_diff == 1)         {         return "$year_diff year";     }         else if ($year_diff > 1)         {         return "$year_diff years";     }     } echo getabsage("2012-02-29") ?> 

also if can suggest better code please update it.

i need find date of birth in months , days if person less 1 year old.

i having latest 5.4 php version on server.

with 2012-02-29, returning 2 years whereas should 3 years. please help.

why not using date_diff() function way through? give desired result:

function getabsage($birthday) {      $age = '';     $diff = date_diff(date_create(), date_create($birthday));     $years = $diff->format("%y");     $months = $diff->format("%m");     $days = $diff->format("%d");      if ($years) {         $age = ($years < 2) ? '1 year' : "$years years";     } else {         $age = '';         if ($months) $age .= ($months < 2) ? '1 month ' : "$months months ";         if ($days) $age .= ($days < 2) ? '1 day' : "$months days";     }     return trim($age); } 

another way calculating time difference in seconds , taking there:

list($year,$month,$day) = explode("-", $birthday); $diff = mktime(0,0,0,date('n'),date('j'),date('y')) - mktime(0,0,0,$month,$day,$year); 

then day consists of 24 hours each 60 minutes each 60 seconds:

$sday = 60 * 60 * 24; 

and calculating years difference be:

$years = floor($diff / (365.2425 * $sday));      

but stick first version presented using date_diff()