How can I convert Hijri date to gregorian in PHP?

Question

I would like convert this date:09/13/1436 (it is Hijiri) to 2015-06-30(it is Gregorian). I tried that:

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


$date = HijriToJD(09, 13, 1436);

echo jdtogregorian($date);

and when i made compile i got 10/7/2014. Someone have any idea??

maybe take a look at http://php.net/manual/de/ref.calendar.php Comment #4 — donald123, Jul 01 '15 at 09:23
[how to convert hijri date to gregorian date using php](http://voidasks.com/how_to_convert_hijri_date_to_gregorian_date_using_php) — Mark Baker, Jul 01 '15 at 09:33
I made convert on website ,so i know it is must be 2015-06-30. So i dont know why i got 10/7/2014. — Khurian, Jul 01 '15 at 09:34
@john-conde - not really a duplicate of that John.... poster is having problems converting dates between __calendars__, not simply formats — Mark Baker, Jul 01 '15 at 16:46

score 8 · Accepted Answer · answered Jul 01 '15 at 09:42

8

Passing 09 as the month is the problem.... a number with a leading zero is treated as octal in PHP. 09 is invalid octal, so it is treated as a 0.

Calling

$r=HijriToJD(9, 13, 1436);

(without the leading zero for the month) should give you a correct result

answered Jul 01 '15 at 09:42

Mark Baker

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Yes!! Thanks you are my Hero. – Khurian Jul 01 '15 at 09:55
1

I couldn't let you get the month wrong at such an important time of year could I? :) – Mark Baker Jul 01 '15 at 09:57

score 1 · Answer 2 · answered Oct 24 '19 at 11:17

More accurate will be

return floor((11 * $year + 3) / 30) + floor(354 * $year) + floor(30 * $month)
            - floor(($month - 1) / 2) + $day + 1948440 - 386;

You can use the floor instead of int

https://github.com/OpenSID/OpenSID/blob/master/donjo-app/libraries/Date_conv.php#L44

How can I convert Hijri date to gregorian in PHP?

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