jdtounix

(PHP 4, PHP 5, PHP 7, PHP 8)

jdtounix — 转变Julian Day计数为一个Unix时间戳

说明

jdtounix(int $jday): int

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）。

参数

jday: 一个在 2440588 到 2465342 之间的julian天数

返回值

指定的julian天数的开始时的时间戳。

参见

unixtojd() - 转变Unix时间戳为Julian Day计数

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User Contributed Notes 5 notes

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fabio at llgp dot org ¶

15 years ago


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

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Anonymous ¶

17 years ago


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!!

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seb at carbonauts dot com ¶

18 years ago


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.

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pipian at pipian dot com ¶

19 years ago


Remember that UNIX timestamps indicate a number of seconds from midnight of January 1, 1970 on the Gregorian calendar, not the Julian Calendar.

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-2

Saeed Hubaishan ¶

7 years ago


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

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