jomali:
Use this:
echo preg_replace('#(\d{2})/(\d{2})/(\d{4})#' , "\\3-\\2-\\1", '24/07/2013'
); RESULT => 2013-07-24
Alejandro M.S
-Mensagem original-
De: jomali [mailto:jomali3...@gmail.com]
Enviada em: sexta-feira, 26 de julho de 2013 17:38
Para: Robert Cummings
Cc: Karl-Arne Gjersøyen; PHP Mailinglist
Assunto: Re: [PHP] From 24/7/2013 to 2013-07-24
On Fri, Jul 26, 2013 at 1:08 PM, Robert Cummings
wrote:
> On 13-07-26 11:42 AM, jomali wrote:
>
>> On Fri, Jul 26, 2013 at 5:18 AM, Karl-Arne Gjersøyen
>> > >wrote:
>>
>> Below is something I try that ofcourse not work because of rsosort.
>>> Here is my code:
>>> ---
>>> $lagret_dato = $_POST['lagret_dato'];
>>> foreach($lagret_dato as $dag){
>>>
>>> $dag = explode("/", $dag);
>>> rsort($dag);
>>> $dag = implode("-", $dag);
>>> var_dump($dag);
>>>
>>> What I want is a way to rewrite contents of a variable like this:
>>>
>>> From 24/7/2013 to 2013-07-24
>>>
>>> Is there a way in PHP to do this?
>>>
>>> Thank you very much.
>>>
>>> Karl
>>>
>>>
>> $conv_date = str_replace('/', '-','24/7/2013'); echo date('Y-m-d',
>> strtotime($conv_date));
>> Result: 2013-07-24
>>
>
> It would be better if you reformatted first since this is ambiguous
> when you have the following date:
>
> 6/7/2013
>
> Here's a completely unambiguous solution:
>
>
> $old = '24/7/2013';
>
> $paddy = function( $bit ){ return str_pad( $bit, 2, '0',
> STR_PAD_LEFT ); };
> $new = implode( '-', array_map( $paddy, array_reverse( explode(
> '/', $old ) ) ) );
>
> echo $new."\n";
>
> ?>
>
> Cheers,
> Rob.
> --
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>
The original question was about reformatting a European (Day/Month/Year)
date. Your solution does not address this problem. Mine assumes the European
date format explicitly.
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