SunshinePHP Developer Conference 2015

GregorianToJD

(PHP 4, PHP 5)

GregorianToJDConverts a Gregorian date to Julian Day Count

Description

int gregoriantojd ( int $month , int $day , int $year )

Valid Range for Gregorian Calendar 4714 B.C. to 9999 A.D.

Although this function can handle dates all the way back to 4714 B.C., such use may not be meaningful. The Gregorian calendar was not instituted until October 15, 1582 (or October 5, 1582 in the Julian calendar). Some countries did not accept it until much later. For example, Britain converted in 1752, The USSR in 1918 and Greece in 1923. Most European countries used the Julian calendar prior to the Gregorian.

Parameters

month

The month as a number from 1 (for January) to 12 (for December)

day

The day as a number from 1 to 31

year

The year as a number between -4714 and 9999

Return Values

The julian day for the given gregorian date as an integer.

Examples

Example #1 Calendar functions

<?php
$jd 
GregorianToJD(10111970);
echo 
"$jd\n";
$gregorian JDToGregorian($jd);
echo 
"$gregorian\n";
?>

See Also

  • jdtogregorian() - Converts Julian Day Count to Gregorian date
  • cal_to_jd() - Converts from a supported calendar to Julian Day Count

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User Contributed Notes 5 notes

up
1
ryker at ridgex dot net
8 years ago
<?php
/*
* ComputeDateDifference(...)
*   Description:
*     Calculates the difference between two dates.
*
*   Parameter:
*     $m0, $d0, $y0   => 1. Moth/Day/Year
*     $m1, $d1, $y1   => 2. Moth/Day/Year
*
*   Return:
*     Difference between given dates in days.
*
*   Autor:
*     06.06.2006 - Christian Meyer <ryker@ridgex.net>
*/
function ComputeDateDifference($m0,$d0,$y0,$m1,$d1,$y1)
{
 
$x0 = gregoriantojd($m0,$d0,$y0);
 
$x1 = gregoriantojd($m1,$d1,$y1); 
 
 
$diff = $x1 - $x0;
 
  if (
$diff < 0)
   
$diff *= -1; // abs
   
 
return $diff;   
}
?>
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1
httpwebwitch
10 years ago
This function also ignores decimal fractions in JD dates, and it uses non-standard format for returning the Gregorian date.

So, if your JD date is 2453056.28673, the Gregorian returned value is 2/20/2004, not "2004-02-20 23:45:36"

The decimal part is important, since the Julian day begins at noon, for example 2453056.49 is on Friday, 2453056.50 is on Saturday. Discarding the decimal part means that your returned Gregorian Date will be wrong 50% of the time.
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0
brian at slaapkop dot net
4 years ago
Some people might find it useful to use a function that takes in dates formatted by ISO8601 standards (yyyy-mm-dd):

<?php function daysBetweenDate($from, $till) {
/*
*This function will calculate the difference between two given dates.
*
*Please input time by ISO 8601 standards (yyyy-mm-dd).
*i.e: daysBetweenDate('2009-01-01', '2010-01-01');
*This will return 365.
*
*Author: brian [at] slaapkop [dot] net
*May 5th 2010
*/
   
if($till < $from) {
       
trigger_error("The date till is before the date from", E_USER_NOTICE);
        }
       
   
//Explode date since gregoriantojd() requires mm, dd, yyyy input;
       
$from = explode('-', $from);
       
$till = explode('-', $till);

   
//Calculate date to Julian Day Count with freshly created array $from.
       
$from = gregoriantojd($from[1], $from[2], $from[0])."<br />";
       
   
//Calculate date to Julian Day Count with freshly created array $till.
       
$till = gregoriantojd($till[1], $till[2], $till[0])."<br />";

   
//Substract the days $till (largest number) from $from (smallest number) to get the amount of days
       
$days = ($till - $from);
   
   
//Return the number of days.
       
return $days;

   
//Isn't it sad how my comments use more lines than the actual code?
}
?>
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0
jfg
5 years ago
If you need the same output as the g_date_get_julian function of the GlibC, here is my php implementation :

<?php
   
/**
     * Glib g_date_get_julian PHP implementation
     *
     * @param  $str  Date string in a format accepted by strtotime
     * @author jfg
     */
   
private function _get_julian( $str )
    {
       
$d = date_create($str);

        if(
$d == false )
            return
0;
       
       
$day_in_year = (int) date_format($d, "z");
       
$year        = (int) date_format($d, "Y") - 1;
       
$julian_days = $year * 365;
       
$julian_days += ($year >>= 2);
       
$julian_days -= ($year /= 25);
       
$julian_days += $year >> 2;
       
$julian_days += $day_in_year + 1;

        return
ceil($julian_days);
    }

?>
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0
jettyrat at jettyfishing dot com
9 years ago
You can obtain the decimal fraction of the Julian date with the php gregoriantojd() function or the function shown below by applying this code to the returned value.

<?php
  $julianDate
= gregoriantojd($month, $day, $year);

 
//correct for half-day offset
 
$dayfrac = date('G') / 24 - .5;
  if (
$dayfrac < 0) $dayfrac += 1;

 
//now set the fraction of a day
 
$frac = $dayfrac + (date('i') + date('s') / 60) / 60 / 24;

 
$julianDate = $julianDate + $frac;
?>
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