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：

<?PHPfunction 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。