#20090 [Opn->Bgs]: Date/POSIX Problem: Two October 27s??
| From: | rasmus@php.net | Date: | Fri, 25 Oct 2002 16:46:51 +0000 |
| Subject: | #20090 [Opn->Bgs]: Date/POSIX Problem: Two October 27s?? | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-23325@lists.php.net to get a copy of this message | ||
ID: 20090
Updated by: rasmus@php.net
Reported By: wleoni75@hotmail.com
-Status: Open
+Status: Bogus
Bug Type: Date/time related
Operating System: Linux
PHP Version: 4.2.3
New Comment:
Congratulations, you have discovered daylight savings time.
Previous Comments:
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[2002-10-25 11:44:26] wleoni75@hotmail.com
Having a problem with date(). I simplified the program I am working on
so you can copy and paste the code and see the problem right away.
I start with the month of october and year of 2002 ($month and $year).
I convert that to the POSIX time ($time) of the the first second of
that month.
Then, I keep adding 24*3600 seconds to that time until the end of the
month.
If you run the program, you will see I get TWO 27/10/2002 -- WHY??
WHY? WHY? WHY?
Here is the code:
================ START CODE =======================
$year = "2002";
$month = "10";
$time = mktime (0,0,0,$month,1,$year);
$last_day_of_the_month = date("t",$time);
echo "Last day of the month ($month/$year):
$last_day_of_the_month<br><br>";
for ($day=1;$day<=$last_day_of_the_month;$day++) {
$new_time = $time + ($day-1)*24*3600;
echo date("m/d/Y",$new_time);
echo " (seconds added to original time: ".(($day-1)*24*3600).")";
echo "<br>";
}
=================== END CODE =====================
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=20090&edit=1