Checking For Leap Year Using PHP

Discussion

1. Keith

   Placing a function call as an argument default will result in a fatal error. Also, give this version a shot:

   function is_leap_year( $year = NULL )
   {
       if_numeric( $year ) || $year = date( 'Y' );
       return checkdate( 2, 29, ( int ) $year );
   }
2. Keith

   Whoops, small typo… if_numeric( $year ) should be is_numeric( $year )… :x
3. Tamlyn

   Better yet, use date('L') which returns 1 if it’s a leap year, 0 if it isn’t.
4. Jasa Web Desain

   I do agree with tamlyn, why you dont use date function??
   It can makes a load fasting right?
5. Travis

   date(‘L’) is way better because leap year is not every 4 years.
6. Indra

   Date("L") only tell you in a given year, default is the year today.
   If you need to know whether previous or next year is a leap, you must reset the date.
   It could affect the system.
• Lossy

  Indra, you can always pass the timestamp as second parameter to date function:

  echo date('L', mktime(1, 1, 1, 1, 1, 2005));
  
7. iCalculator

   I think this is a better example relating to the function. Also provided example for anyone looking for true/false as I was in this instance.

   /* for true or false */
   function is_leap_year($year)
   {
   	return ( date ('L', mktime(1,1,1,1,1, $year) ) === 1 ) ? true : false;
   }
   /*for 0 or 1 Whether it's a leap year: 1 if it is a leap year, 0 otherwise. */
   function is_leap_year($year)
   {
   	return date ('L', mktime(1,1,1,1,1, $year) );		
   }
   
8. Habibur Rahaman

   <?php
   $day = "";
   for($i=0; $i<4; $i++)
   {
       $day =  date("d", mktime(0, 0, 0, 2, 29, date("Y")+$i));
       if($day == 29)
       {
           $year = date("Y")+$i;
           break;
       }
   }
   echo "The next leap year is 29th February $year";    
   ?>
   
   
9. Milan Dvořák

   code of Habibur Rahaman will not work for example for the year 1897, because 1900 is not leap year.