php - Convert to date format dd/mm/yyyy

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Learn more about Teams I find on a lot of our servers (operating on Australian time) I need to use: date("d/m/o", strtotime($str)); -> i.e 'o' for the year. – niggles Apr 20, 2010 at 1:57 Hi there! I have already tried your solution but does not seems to work. Like I asked on my post, I just need the date in the dd/mm/yyyy format without the time component. Thanks for the help, Best regards! – Rui Gonçalves Apr 20, 2010 at 9:09 Which part doesn't work?


    strtotime("2010-04-19 18:31:27");

returns the seconds since the epoch,

1271716287

(for my TZ).


    date()

takes a format string and an epoch int; the format string tells it what to return, and "d/m/Y" is days as a zero-padded int, months as a zero-padded int, and year as four digits. Since no time parameters are included in the format string (e.g. "d/m/Y h:i:s"), they won't be returned – Michael Mrozek Apr 20, 2010 at 14:03 Hi there! I was commiting an error with the application of the suggested solution. It worked just fine! To all others, thanks for the responses. Thanks again for the response! Best regards – Rui Gonçalves Apr 21, 2010 at 21:52

If your date is in the format of a string use the explode function

    array explode ( string $delimiter , string $string [, int $limit ] )
//In the case of your code
$length = strrpos($oldDate," ");
$newDate = explode( "-" , substr($oldDate,$length));
$output = $newDate[2]."/".$newDate[1]."/".$newDate[0];
Hope the above works now 
Simply Try
$date=date('d/m/Y',strtotime($yourdatevariable))
or else you want to convert to d-m-Y
Just Simply Edit above d/m/Y with d-m-Y
$source    =    'your varible name';
$date    =     new DateTime($source);
$_REQUEST["date"]    =     $date->format('d-m-Y');
echo $_REQUEST["date"];
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