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# GregorianToJD

(PHP 4, PHP 5, PHP 7)

GregorianToJD转变一个Gregorian历法日期到Julian Day计数

### 说明

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

Gregorian历法的合理范围是4714 B.C. 至 9999 A.D.

`month`

`day`

`year`

### 范例

Example #1 Calendar functions

``` <?php\$jd = GregorianToJD(10, 11, 1970);echo "\$jd\n";\$gregorian = JDToGregorian(\$jd);echo "\$gregorian\n";?> ```

### 参见

• jdtogregorian() - 转变一个Julian Day计数为Gregorian历法日期
• cal_to_jd() - 从一个支持的历法转变为Julian Day计数。

### User Contributed Notes 6 notes

httpwebwitch
13 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. ```
ryker at ridgex dot net
11 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;    } ?> ```
brian at slaapkop dot net
7 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? }?> ```
jfg
8 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);    }?> ```
jettyrat at jettyfishing dot com
12 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; ?> ```
-1
klauspeter
7 months ago
``` Wie kann ich mir das erklären?Bei der Eingaben von gregoriantojd (1 , 1, -4714);  // erhalte ich 0, das ist OK, 01.01.4713 BCgregoriantojd (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 ?) ```