i've got code
$dss='2011 04 28 10:00:45';
echo date('Y', strtotime($dss));
it display 1970- why?
-
possible duplicate of strtotime() & date() weired behaviour when converting date in to same format as it was beforeShakti Singh– Shakti Singh2011-04-28 07:28:10 +00:00Commented Apr 28, 2011 at 7:28
Add a comment
|
4 Answers
it display 1970- why?
Becuase the format YYYY MM DD HH:MM:SS isn't one of the formats that strtotime accepts. In particular, it's failing to find a valid date string.
If you're using PHP 5.3 or better, look at DateTime::createFromFormat(), which takes a date-formatted string and creates a new DateTime object based on that format.
If you're stuck on a lower version of PHP, your "best" choice is the clunky strptime.
Or you can just stick dashes between the YYYY and MM and then between the MM and the DD, yielding YYYY-MM-DD, a perfectly cromulent format.
As Charles says it's not a valid date format. This should reformat it:
$dss = preg_replace('/^(\d+) (\d+) (.+)$/i', '$1-$2-$3', $dss);
Comments
It seems that you want only the year from this sting. You can simply do this:
$dss='2011 04 28 10:00:45';
echo substr($dss, 0, 5);
Comments
Wrong answer before...was -1 worthy...I'm now climbing backwards onto the donkey, putting the pail on my head, and leaving town.
5 Comments
Ben
Where did the 'Delete' link go? Read too quickly and now regret it :/
Charles
@Steve, I can't find one for my own answer either. Weird. Maybe you can edit your answer into correctness? (Hint: On error,
strtotime returns either false or -1 based on the PHP version being used.)Ben
Yes, was, and is, Weird. Thanks for the tip though! Taken, with some self-deprecating flippancy.
Charles
@Steve, I filed a bug, looks like it'll be fixed very soon (moments?)
Ben
Great, thanks. And to the -1 person, cheers on the reversal :)