PHP method to calculate the accurate date of Easter

This article mainly introduces the method of accurately calculating the Easter date in PHP. It involves the skills of PHP operating dates. It is of great practical value. Friends in need can refer to it.

The example in this article tells the accurate calculation of Easter in PHP. Festival method. Share it with everyone for your reference. The details are as follows:

<?PHP
function isLeapYear( $nYEAR ) {
 if((($nYEAR % 4 == 0) AND !($nYEAR % 100 == 0)) AND ($nYEAR % 400 != 0))
 {
  return TRUE;
 } else {
  return FALSE;
 }
}
function p( $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 = p( $nYEAR, 100 ) + 1;
  $nKSJ = p( 3 * $nJHD, 4 ) - 12;
  $nKORR = p( 8 * $nJHD + 5, 25 ) - 5;
  $nSO = p( 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 = p( 20 * $nNM, 611 ) - 2;
  // Day
  $nNT = $nN;
  $nNT = $nN;
  if ( $nNT > ( 59 + $nA )) {
   $nNT = $nNT + 2 - $nA;
  }
  $nNT = $nNT + 91;
  $nM = p( 20 * $nNT, 611 );
  $nDAY = $nNT - p( 611 * $nM, 20 );
  $dtEasterSunday = mktime( 0,0,0,$nMONTH,$nDAY,$nYEAR );
 }
 return $dtEasterSunday;
}
?>

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