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。

php准确计算复活节日期的方法

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