PHP 7.1.21 Released

jdtounix

(PHP 4, PHP 5, PHP 7)

jdtounix转变Julian Day计数为一个Unix时间戳

说明

int jdtounix ( int $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)。

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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3
fabio at llgp dot org
11 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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2
seb at carbonauts dot com
14 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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2
Anonymous
13 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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0
Saeed Hubaishan
3 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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0
pipian at pipian dot com
15 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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