这篇文章主要介绍了php准确计算复活节日期的方法,涉及php操作日期的技巧,非常具有实用价值,需要的朋友可以参考下
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。
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