How do I convert the date format from 2012-03-19T22:10:56.000Z to Wed Nov 13 18:06:37 +0000 2013 without using string functions like substr. This date format is being used by Gnip. Not sure what format it is.
I tried using the date function
$postedTime = "2012-03-19T22:10:56.000Z";
date($postedTime,"D M d H:i:s O Y");
But I get an error since $postedTime is a string and not long variable.
Why use substr() when you can use functions designed specifically for working with dates?
$date = new DateTime('2012-03-19T22:10:56.000Z');
echo $date->format('r Y');
You need to swap the order in your date function
date('format string', $time)
So you want:
date("D M d H:i:s O Y", strtotime($postedTime));
You can transform the string in a DateTime object with
$date = new DateTime($postedTime);
Then you may need to change the time zone (since your postedTime is timezone Z for Zulu)
$date->setTimezone(new DateTimeZone('Europe/Rome'));
and maybe also add or subtract a suitable offset DateInterval.
Finally, if you need a timestamp for use in date, you do so with [getTimeStamp][1].
$ts = $date->getTimeStamp();
If you have all datetimes in Zulu, just use
$date = new DateTime($postedTime);
date("Y-m-d H:i:s", $date->getTimeStamp());
or if you do not need the timestamp,
$date = new DateTime($postedTime);
print $date->format('Y-m-d H:i:s');
You should get the same date you entered. If it is offset one or more hours plus or minus, you have a time zone problem.
