ID: 9076
Updated by: jmoore
Reported By: [EMAIL PROTECTED]
Old-Status: Assigned
Status: Closed
Bug Type: Calendar related
Operating system:
PHP Version: 4.0.4pl1
Assigned To:
Comments:
I cant reproduce this at all..
The following script:
<?php
$jd = GregorianToJD (02,02,2001);
echo "$jd\n";
$gregorian = JDToGregorian ($jd);
echo "$gregorian\n";
$jd = GregorianToJD (7,6,1980);
echo "$jd\n";
$gregorian = JDToGregorian ($jd);
echo "$gregorian\n";
?>
Gives this output..
D:\cvs\php4\Release_TSDbg>php -q test.php
2451943
2/2/2001
2444427
7/6/1980
which is what I would expect.
Please reopen bug report if I am wrong (and point out why I am wrong). Also if your
system still displays incorrect dates with latest cvs or 4.0.6RC1 please reopen too
giveing more information about your system.
- James
Previous Comments:
---------------------------------------------------------------------------
[2001-04-29 05:50:00] [EMAIL PROTECTED]
Ill look at these patches at some point soon.
- James
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[2001-02-09 07:15:34] [EMAIL PROTECTED]
I found more thoroughly-researched algorithms for
calculating to and from Gregorian/Julian calender dates
to Julian Day Count values:
http://www.capecod.net/~pbaum/date/date0.htm
I suggest that the PHP Calender functions be based
upon the algorithms here, as meticulous care seems
to have been taken to formulate and proof them.
In the meantime, I have written my own
implementations of the GregorianToJD and
JDToGregorian functions in PHP. I have tested them
using the examples in Table 2 of Chapter 1 of Baum's
work and they pass.
function php_gregoriantojd($input_year, $input_month,
$input_day) {
// Make sure supplied arguments are of the correct
form. Day may be a fractional value,
// but month and year must be integers.
$input_year = intval($input_year);
$input_month = intval($input_month);
$input_day = doubleval($input_day);
// Adjust the start of the year so that it is in March.
if($input_month < 3) {
$input_month += 12;
$input_year -= 1;
}
// Calculate and return the Julian Day Count.
return $input_day + floor((153 * $input_month - 457) /
5) + 365 * $input_year + floor($input_year / 4) -
floor($input_year / 100) + floor($input_year / 400) +
1721118.5;
}
function php_jdtogregorian($jdc) {
// Make sure that the Julian Day Count value is proper.
$jdc = doubleval($jdc);
$z = floor($jdc - 1721118.5);
$r = $jdc - 1721118.5 - $z;
$g = $z - 0.25;
$a = floor($g / 36524.25);
$b = $a - floor($a / 4);
$gregorian["year"] = floor(($b + $g) / 365.25);
$c = $b + $z - floor(365.25 * $gregorian["year"]);
$gregorian["month"] = floor((5 * $c + 456) / 153);
$gregorian["day"] = $c - floor((153 *
$gregorian["month"] - 457) / 5) + $r;
if($gregorian["month"] > 12) {
$gregorian["year"] += 1;
$gregorian["month"] -= 12;
}
return $gregorian;
}
The line breaks on all of that will probably be all
mangled when I post, but it should make sense when
straightened out...
---------------------------------------------------------------------------
[2001-02-02 11:05:26] [EMAIL PROTECTED]
The GregorianToJD Calender function returns incorrect
results. As far as I can tell, this applies to the
GregorianToSdn internal function in the PHP source.
There's not really any point putting a sample PHP script
here, as it's the algorithm that seems to be at fault. I
took two specific cases and worked through the
algorithm by hand (i.e. pen and paper). Here are the
results:
2nd February 2001 (2/2/2001) = 2451943
7th June 1980 (7/6/1980) = 2473618
As I understand it, the Julian Day Count is a uniform
count of days from some time around 4714 BC. Now,
how can 2/2/01 be a lesser number of days than
7/6/80?
I have no idea what the correct results for these test
cases are, nor any idea what particular part of the
algorithm is wrong. It's just wrong somewhere!
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