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
(PHP 4, PHP 5, PHP 7)
jdtounix — 转变Julian Day计数为一个Unix时间戳
$jday
)
这个函数根据给定的julian天数返回一个Unix时间戳，或如果参数jday不在Unix时间（Gregorian历法的1970年至2037年，或2440588 <= jday <= 2465342）范围内返回 FALSE 。返回的时间是本地时间（不是GMT）。
jday一个在 2440588 到 2465342 之间的julian天数
指定的julian天数的开始时的时间戳。
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
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
Warning: 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.