php准确计算复活节日期的方法
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2024-01-29 20:03:52
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：
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。