JDToGregorian

(PHP 4, PHP 5, PHP 7, PHP 8)

JDToGregorian — Converte il Giorno Giuliano in data Gregoriana

Descrizione

jdtogregorian(int $giornogiuliano): string

Converte il Giorno Giuliano in una stringa contenente la data Gregoriana nel formato "mese/giorno/anno".

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User Contributed Notes 5 notes

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httpwebwitch ¶

20 years ago

JD days may have decimal fractions which correspond to the time of day. The Julian day begins at noon, and the decimal fraction measures fractional days until the beginning of the next day at noon.

For instance, Julian Day 2453179.00000 is June 22, 2004, at 12:00pm (noon). 

One hour later, it's 2453179.04167
At 2453179.20833 I'll have dinner, and
at 2453179.45833, it's time for the evening news.
After a good night of sleep, my alarm will go off at 2453179.83333, 
then at noon on June 23, a new Julian Day begins at 2453180.

To use these functions with fractional days, strip the fractional part with floor(), and apply the function to the integer part. 

Then add 12 hours, bringing you to noon of that day. That is the actual time returned by JDToGregorian().

Then add the fractional part of the day, by multiplying the decimal part of the Julian Day by (24*60*60) seconds. This may take you forward or backward to a different Gregorian calendar date.

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treebe ¶

21 years ago

Julian to Gregorian date change.

If you do not have the calendar extensions loaded this is little function works realy well.



<?php

function jd_to_greg($julian) {

    $julian = $julian - 1721119;

    $calc1 = 4 * $julian - 1;

    $year = floor($calc1 / 146097);

    $julian = floor($calc1 - 146097 * $year);

    $day = floor($julian / 4);

    $calc2 = 4 * $day + 3;

    $julian = floor($calc2 / 1461);

    $day = $calc2 - 1461 * $julian;

    $day = floor(($day + 4) / 4);

    $calc3 = 5 * $day - 3;

    $month = floor($calc3 / 153);

    $day = $calc3 - 153 * $month;

    $day = floor(($day + 5) / 5);

    $year = 100 * $year + $julian;



    if ($month < 10) {

        $month = $month + 3;

    }

    else {

        $month = $month - 9;

        $year = $year + 1;

    }

    return "$day.$month.$year";

}

?>

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swiles at ddbc dot edu dot tw ¶

15 years ago

The php gregoriantojd() and jdtogregorian() functions, in addition to the limitations noted by httpwebwitch, does not take account of the 'Astronomical' system of reckoning - i.e. using a year zero, instead of going directly from 1BCE to 1CE, as with the Christian Anno Domini system.

These functions can be used to wrap the php built-ins to return ISO 8601 compliant dates:

<?php
function ISO8601toJD($ceDate) {
    list($day, $month, $year) = array_map('strrev',explode('-', strrev($ceDate), 3));
    if ($year <= 0) $year--;
    return gregoriantojd($month, $day, $year);
}

function JDtoISO8601($JD) {
    if ($JD <= 1721425) $JD += 365;
    list($month, $day, $year) = explode('/', jdtogregorian($JD));
    return sprintf('%+05d-%02d-%02d', $year, $month, $day);
}
?>

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ashton at ieee dot org ¶

8 years ago

The minimum acceptable input Julian day count is 1, which produces an output of "11/25/-4714" (at least in my operating system and location) which means 25 November 4714 BC. PHP does not recognize the year 0. Astronomers do use the year 0, and would write the Gregorian date that corresponds to Julian day number 1 as 25 November -4713.

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-1

uni_fl4r3 at t hotmail dot com ¶

21 years ago

I have made a slight modification to treebe's jd to greg function, this one will transform a unix timestamp to Gregorian day/month/year format...



<?php

function unix_to_greg($unix_timestamp) {

    $julian = floor(((($unix_timestamp / "60") / "60") / "24") + "2440588");

    $julian = $julian - 1721119;

    $calc1 = 4 * $julian - 1;

    $year = floor($calc1 / 146097);

    $julian = floor($calc1 - 146097 * $year);

    $day = floor($julian / 4);

    $calc2 = 4 * $day + 3;

    $julian = floor($calc2 / 1461);

    $day = $calc2 - 1461 * $julian;

    $day = floor(($day + 4) / 4);

    $calc3 = 5 * $day - 3;

    $month = floor($calc3 / 153);

    $day = $calc3 - 153 * $month;

    $day = floor(($day + 5) / 5);

    $year = 100 * $year + $julian;

    

    if ($month < 10)

    {

        $month = $month + 3;

    }else{

        $month = $month - 9;

        $year = $year + 1;

    }

    return "$day.$month.$year";

}

?>

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