php date strtotime wrong unix tuimestamp - php

Question

I am tryig to check timestamp using php. However, the $date variable echoes out a timestamp
1382389873 which I checked with a unix converter epochconverter.com it shows Mon, 21 Oct 2013 21:11:13 GMT when it is meant to be 5 Nov 2013. Anyone can see the code below and pinpount my mistake? Thanks.
$today = strtotime(date("d.m.y"));
$c_date = strtotime($CI->session->userdata('c_date'));
$early = strtotime(date("d.m.y")."+2 week");
$date = strtotime("5.11.13");
echo'<br/>';
echo 'test'.$date;
echo'<br/>';
echo 'collection'.$c_date;
echo'<br/>';
echo 'early'.$early;
echo'<br/>';
echo 'today '.$today;
echo'<br/>';

Answer 1

strtotime is very vague in the way it processes dates. It is interpreting "5.11.13" as 5:11:13pm today (Which is 21:11:13 on a 24-hour clock).
If you want to specify november 5th you should do it like so:
$date = strtotime("11/5/13");
echo $date;
echo date("m/d/Y", $date);
Output:
1383609600
11/05/2013

Answer 2

Others answered this before I did, but here are my tests and notes to supplement the conversation... strtotime() converts a string to a time integer. date() converts an integer to a (optionally formatted) string. I was gonna say you can always just use that rather than epochvonverter.com... but now I understand why you went there. Also note that the manual says this:
"Note: Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed.
To avoid potential ambiguity, it's best to use ISO 8601 (YYYY-MM-DD) dates or DateTime::createFromFormat() when possible." -http://www.php.net/manual/en/function.strtotime.php (more quirky things on that page, btw).
It is very interesting that this example with dot is NOT producing d.m.y, but h.m.s! I guess that's because we're working with strtoTIME, not... uh.. strtoDATE :)
Consider this:
$integer = 1382389873;
var_dump($integer);
echo "<br>";
//int(1382389873)
$string = date($integer);
var_dump($string);
echo "<br>";
//string(10) "1382389873"
$formatted=date('d.m.y',$integer);
$laterint=strtotime($formatted."+2 week");
var_dump($laterint);
echo "<br>";
// int(1383682213)
$laterstring=date('d.m.y',$laterint);
var_dump($laterstring);
echo "<br>";
// string(8) "05.11.13"
And here's the juicy part:
$date1 = strtotime("5.11.13");
var_dump($date1);
echo "<br>";
$datestring1=date('d.m.y',$date1);
var_dump($datestring1);
//BAD PHP, BAD!
echo "<br>";
$date2 = strtotime("5.11.2013");
var_dump($date2);
echo "<br>";
$datestring2=date('d.m.y',$date2);
var_dump($datestring2);
echo "<br>";
$date3 = strtotime("5/11/13");
var_dump($date3);
echo "<br>";
$datestring3=date('d.m.y',$date3);
var_dump($datestring3);
echo "<br>";
Really interesting stuff, guys - thank you all! The moral of the story for me is to always be explicit with 4 digit year.

Answer 3

I ran a couple of tests for you and it seems that you need to put your year in YYYY format e.g. 2013:
$date = strtotime("5.11.2013"); // your code
echo $date . "\n";
echo date('d-m-Y', $date);
// timestamp: 1383562800
// converted to date: 05-11-2013
--edit-- as I mentioned in my previous comment, if $date is supposed to be the same as $early, why are you even bothering to reassign it manually? Why not just use $early?