If you need an easy way to convert a decimal julian day to an unix timestamp you can use:
$unixTimeStamp = ($julianDay - 2440587.5) * 86400;
2440587.5 is the julian day at 1/1/1970 0:00 UTC
86400 is the number of seconds in a day
(PHP 4, PHP 5, PHP 7, PHP 8)
jdtounix — 将儒略日数转换为 Unix 时间戳
$julian_day): int
这个函数将返回与 julian_day
给出的儒略日相对应的 Unix 时间戳。返回的时间是 UTC。
julian_day
儒略日数在 64 位系统上是 2440588 到 106751993607888
之间的数字，在 32 位系统上是 2440588 到 2465443 之间的数字。
指定儒略日的开始时（午夜，而非中午）的 unix 时间戳。
julian_day 超出允许的范围值，抛出 ValueError。
| 版本 | 说明 |
|---|---|
| 8.0.0 |
此函数在失败时不再返回 false，而是引发 ValueError。
|
| 7.3.24, 7.4.12 |
扩充了 julian_day 的上限。以前无论计算机体系结构如何，都是
2465342。
|
If you need an easy way to convert a decimal julian day to an unix timestamp you can use:
$unixTimeStamp = ($julianDay - 2440587.5) * 86400;
2440587.5 is the julian day at 1/1/1970 0:00 UTC
86400 is the number of seconds in a dayWarning: the calender functions involving julian day operations seem to ignore the decimal part of the julian day count.
This means that the returned date is wrong 50% of the time, since a julian day starts at decimal .5 . Take care!!Remember that unixtojd() assumes your timestamp is in GMT, but jdtounix() returns a timestamp in localtime.
This fooled me a few times.
So if you have:
$timestamp1 = time();
$timestamp2 = jdtounix(unixtojd($timestamp1));
Unless your localtime is the same as GMT, $timestamp1 will not equal $timestamp2.Remember that UNIX timestamps indicate a number of seconds from midnight of January 1, 1970 on the Gregorian calendar, not the Julian Calendar.unixtojd() assumes that your timestamp is in GMT, but jdtounix() returns a timestamp in localtime.
so
<?php
$d1=jdtogregorian(unixtojd(time()));
$d2= gmdate("m/d/Y");
$d3=date("m/d/Y");
?>
$d1 always equals $d2 but $d1 may differ from $d3