javascript - 如何使用 PHP 将回历日期转换为公历日期？

Question

请帮帮我

如何使用 php 将回历日期转换为公历日期

如果您知道任何可以将回历转换为公历的日期选择器，请告诉我。

Answer 1

检查这个solution .在这里引用

<?php

class HijriCalendar
{
    function monthName($i) // $i = 1..12
    {
        static $month  = array(
            "Muḥarram", "Safar", "Rabī‘ I", "Rabī‘ II",
            "Jumādá I", "Jumādá II", "Rajab", "Sha‘bān",
            "Ramaḍān", "Shawwāl", "Dhū al-Qa‘dah", "Dhū al-Ḥijjah"
        );
        return $month[$i-1];
    }

    function GregorianToHijri($time = null)
    {
        if ($time === null) $time = time();
        $m = date('m', $time);
        $d = date('d', $time);
        $y = date('Y', $time);

        return HijriCalendar::JDToHijri(
            cal_to_jd(CAL_GREGORIAN, $m, $d, $y));
    }

    function HijriToGregorian($m, $d, $y)
    {
        return jd_to_cal(CAL_GREGORIAN,
            HijriCalendar::HijriToJD($m, $d, $y));
    }

    # Julian Day Count To Hijri
    function JDToHijri($jd)
    {
        $jd = $jd - 1948440 + 10632;
        $n  = (int)(($jd - 1) / 10631);
        $jd = $jd - 10631 * $n + 354;
        $j  = ((int)((10985 - $jd) / 5316)) *
            ((int)(50 * $jd / 17719)) +
            ((int)($jd / 5670)) *
            ((int)(43 * $jd / 15238));
        $jd = $jd - ((int)((30 - $j) / 15)) *
            ((int)((17719 * $j) / 50)) -
            ((int)($j / 16)) *
            ((int)((15238 * $j) / 43)) + 29;
        $m  = (int)(24 * $jd / 709);
        $d  = $jd - (int)(709 * $m / 24);
        $y  = 30*$n + $j - 30;

        return array($m, $d, $y);
    }

    # Hijri To Julian Day Count
    function HijriToJD($m, $d, $y)
    {
        return (int)((11 * $y + 3) / 30) +
            354 * $y + 30 * $m -
            (int)(($m - 1) / 2) + $d + 1948440 - 385;
    }
};

$hijri = HijriCalendar::GregorianToHijri( time() );
echo $hijri[1].'. '.HijriCalendar::monthName($hijri[0]).' '.$hijri[2];

?>