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。
