php date format converting problem

Question

Here my function that i try to trasform dates into different formats.

/*  example:
*   dateString          =   '03/25/2010';
*   inputDateFormat     =   '%m/%d/%Y';
*   ouputDateFormat     =   'Y-m-d';
*   return              ->  '2010-03-25';
*/  
function formatDate($dateString,$inputFormat=NULL,$outputFormat=NULL){
    if($dateString==''||$dateString==NULL) return '';
    $t =  strptime($dateString,$inputFormat);
    return gmdate($outputFormat,mktime($t[tm_sec],$t[tm_min],$t[tm_hour],($t[tm_mon]+1),($t[tm_mday]+1),($t[tm_year]+1900)));
}

My problem is

when i try to convert this date Wed, 19 Jan 2011 21:16:37 +0000 into 2011-01-19 21:16:37 with the following line:

echo formatDate('Wed, 19 Jan 2011 21:16:37 +0000','%a, %d %b %Y %H:%M:%S','Y-m-d H:i:s');

the result is this:

2011-01-21 11:16:21

why i'm getting the date with 2 days extra. Do you have any idea?

Book Of Zeus · Accepted Answer · 2011-08-27 23:36:09Z

5

use this instead:

  function formatDate($dateString, $outputFormat=NULL){
      return date($outputFormat, strtotime($dateString));
  }

  echo formatDate('Wed, 19 Jan 2011 21:16:37 +0000','Y-m-d H:i:s');

answered Aug 27, 2011 at 23:36

Book Of Zeus

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4 Comments

Yunus Eren Güzel Over a year ago

it works thank you :) but i need to specify the input format, too what do you suggest for this?

Book Of Zeus Over a year ago

no problem, as a reminder, you should add validation in case you have an invalid date. before php 5.1 strtotime return -1, after it return FALSE.

Yunus Eren Güzel Over a year ago

when i give such date 03/02/2011 it cannot understand which one is day and which one is month :)

Book Of Zeus Over a year ago

you can even put these: 'now', 'tomorrow' or 'now + 2 days' when you call formatDate

sman591 · Accepted Answer · 2011-08-27 23:36:02Z

4

This is a wild guess, but maybe you need to set yoru time zone?

date_default_timezone_set() (requires PHP 5)

answered Aug 27, 2011 at 23:36

sman591

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