php准确计算复活节日期的方法

本文实例讲述了php准确计算复活节日期的方法。分享给大家供大家参考。具体如下：

    <?PHP
    function isLeapYear( $nYEAR ) {
     if((($nYEAR % 4 == 0) AND !($nYEAR % 100 == 0)) AND ($nYEAR % 400 != 0))
     {
      return TRUE;
     } else {
      return FALSE;
     }
    }
    function div( $a, $b ){
     return( $a - ( $a % $b )) / $b;
    }
    function easterSunday( $nYEAR ) {
     // The function is able to calculate the date 
     //of eastersunday back to the year 325,
     // but mktime() starts at 1970-01-01!
     if ( $nYEAR < 1970 ) {
      $dtEasterSunday = mktime( 1,1,1,1,1,1970 );
     } else {
      $nGZ = ( $nYEAR % 19 ) + 1;
      $nJHD = div( $nYEAR, 100 ) + 1;
      $nKSJ = div( 3 * $nJHD, 4 ) - 12;
      $nKORR = div( 8 * $nJHD + 5, 25 ) - 5;
      $nSO = div( 5 * $nYEAR, 4 ) - $nKSJ - 10;
      $nEPAKTE = (( 11 * $nGZ + 20 + $nKORR - $nKSJ ) % 30 );
      if (( $nEPAKTE == 25 OR $nGZ == 11 ) AND $nEPAKTE == 24 ) {
       $nEPAKTE = $nEPAKTE + 1;
      }
      $nN = 44 - $nEPAKTE;
      if( $nN < 21 ) {
       $nN = $nN + 30;
      }
      $nN = $nN + 7 - (( $nSO + $nN ) % 7 );
      $nN = $nN + isLeapYear( $nYEAR );
      $nN = $nN + 59;
      $nA = isLeapYear( $nYEAR );
      // Month
      $nNM = $nN;
      if ( $nNM > ( 59 + $nA )) {
       $nNM = $nNM + 2 - $nA;
      }
      $nNM = $nNM + 91;
      $nMONTH = div( 20 * $nNM, 611 ) - 2;
      // Day
      $nNT = $nN;
      $nNT = $nN;
      if ( $nNT > ( 59 + $nA )) {
       $nNT = $nNT + 2 - $nA;
      }
      $nNT = $nNT + 91;
      $nM = div( 20 * $nNT, 611 );
      $nDAY = $nNT - div( 611 * $nM, 20 );
      $dtEasterSunday = mktime( 0,0,0,$nMONTH,$nDAY,$nYEAR );
     }
     return $dtEasterSunday;
    }
    ?>

希望本文所述对大家的php程序设计有所帮助。