(PHP 4, PHP 5, PHP 7)
GregorianToJD — 转变一个Gregorian历法日期到Julian Day计数
$month
, int $day
, int $year
)Gregorian历法的合理范围是4714 B.C. 至 9999 A.D.
虽然这个函数可以处理4714 B.C.以前的日期，但是没有意义。Gregorian历法直到1582年10年15日（或是Julian历法的1582年10月5日）才被发明，很久以后一些国家也没有接受它。比如，英国是在1752年开始使用Gregorian历法，苏联是在1918年，希腊是在1923年，大部分的欧洲国家使用Julian历法。
month月份的范围是 1（January）到 12（December）。
day日期的范围是 1到 31。
year年份的范围是 -4714 到 9999。
给定gregorian历法日期的julian天数。
Example #1 Calendar functions
<?php
$jd = GregorianToJD(10, 11, 1970);
echo "$jd\n";
$gregorian = JDToGregorian($jd);
echo "$gregorian\n";
?>
Wie kann ich mir das erkl?ren?
Bei der Eingaben von
gregoriantojd (1 , 1, -4714); // erhalte ich 0, das ist OK, 01.01.4713 BC
gregoriantojd (1 , 1, -4713); // erhalte ich 38 (0 + 365 ????)
gregoriantojd (1 , 1, -4712); // erhalte ich 404 (0 + 730 ????)
gregoriantojd (1 , 1, 0); // erhalte ich 0 (fencepost error ?)
Some people might find it useful to use a function that takes in dates formatted by ISO8601 standards (yyyy-mm-dd):
<?php function daysBetweenDate($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?
}
?>
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);
}
?>
<?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;
}
?>
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;
?>
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.