rszeus wrote:
> No, sory, my bad typing. It's not the problem..
> I have the same on both caxses, only chnage the variable $id
> $file = "screen/temp/7a45gfdi6icpan1jtb1j99o925_1_mini.jpg"
> $id = '70';
> echo preg_replace('#(screen/)temp/(.+?)_1(.+?\.jpg)#', '$1'.$id, $file);
> Get: 0
>
> file = "screen/temp/7a45gfdi6icpan1jtb1j99o925_1_mini.jpg";
> $id = test;
Well, here you are trying to use a constant called test to set your
variable $id. I would try surrounding test with single or double quotes
> echo preg_replace('#(screen/)temp/(.+?)_1(.+?\.jpg)#', '$1'.$id, $file);
>
> Get: screen/test
>
> What is wrong with having na integer on the var ?
>
> Thank you
>
> -----Mensagem original-----
> De: Jim Lucas [mailto:[email protected]]
> Enviada: quarta-feira, 22 de Julho de 2009 18:44
> Para: rszeus
> Cc: 'Kyle Smith'; 'Eddie Drapkin'; [email protected];
> [email protected]
> Assunto: Re: [PHP] Replace in a string with regex
>
> rszeus wrote:
>> Thank you. I undestand now.
>>
>> Anh it’s already workyng the replacemente with letteres. Bu if the variabele
>> is a number it doens’t work. Any ideas ?
>>
>>
>>
>> $file = "screen/temp/7a45gfdi6icpan1jtb1j99o925_1_mini.jpg";
>>
>>
>>
>> $id = '70';
>>
>> echo preg_replace('#(Iscreen/)temp/(.+?)_1(.+?\.jpg)#', '$1'.$id, $file);
>>
>
> You have a great big capital I in there... Try removing that and see
> what happens.
>
>
>> I get 0
>>
>>
>>
>>
>>
>> $id = 'test';
>>
>> echo preg_replace('#(screen/)temp/(.+?)_1(.+?\.jpg)#', '$1'.$id, $file);
>>
>> I get screen/test
>>
>>
>>
>> Any ideas ?
>>
>>
>>
>> Thank you
>>
>>
>>
>> De: Kyle Smith [mailto:[email protected]]
>> Enviada: quarta-feira, 22 de Julho de 2009 17:22
>> Para: rszeus
>> Cc: 'Eddie Drapkin'; [email protected]; 'Jim Lucas';
>> [email protected]
>> Assunto: Re: [PHP] Replace in a string with regex
>>
>>
>>
>> The first match inside () is assigned to $1, the second is assigned to $2,
>> and so on.
>>
>> rszeus wrote:
>>
>> Hi,
>> It doens't work.
>> I get 0_main.jpg if I do that..
>> I don't undestand the point of $1 $2 and $3..
>> In preg_replace('#(screens/)temp/(.+?)_1(_main\.jpg)#', what will be $1 and
>> $2 and $3 ?
>> $ " I already knwo it's the (.+?) but the others didnt' get it.
>>
>> Thank you...
>>
>> -----Mensagem original-----
>> De: Eddie Drapkin [mailto:[email protected]]
>> Enviada: quarta-feira, 22 de Julho de 2009 16:03
>> Para: [email protected]
>> Cc: Jim Lucas; rszeus; [email protected]
>> Assunto: Re: [PHP] Replace in a string with regex
>>
>> On Wed, Jul 22, 2009 at 11:01 AM, Ashley
>> Sheridan <mailto:[email protected]> <[email protected]>
>> wrote:
>>
>>
>> On Wed, 2009-07-22 at 07:54 -0700, Jim Lucas wrote:
>>
>>
>> rszeus wrote:
>>
>>
>> Thank you.
>>
>> What about instead test i want to insert a variable ?
>> Like
>> $id = "30";
>> $file = "screens/temp/7a45gfdi6icpan1jtb1j99o925_1_main.jpg";
>> echo preg_replace('#(screens/)temp/(.+?)_1(_main\.jpg)#', '$1$id$3', $file);
>>
>>
>> Sure that can be done. But you will need to change the second argument
>> to have double quotes so it will be parsed by PHP.
>>
>> Then I would surround YOUR variable with curly brackets to separate it
>> from the rest.
>>
>> echo preg_replace('#(screens/)temp/.+?_1(_main\.jpg)#',
>> "$1{$id}$3",
>> $file);
>>
>>
>>
>> I am confusing " and '.
>>
>> Thank you
>>
>>
>> -----Mensagem original-----
>> De: Eddie Drapkin [mailto:[email protected]]
>> Enviada: quarta-feira, 22 de Julho de 2009 14:12
>> Para: rszeus
>> Cc: [email protected]
>> Assunto: Re: [PHP] Replace in a string with regex
>>
>> On Wed, Jul 22, 2009 at 9:07 AM, rszeus <mailto:[email protected]>
>> <[email protected]> wrote:
>>
>>
>> Hi. It Works to remove the _1 but it doesn't replace
>> '7a45gfdi6icpan1jtb1j99o925' for 'test'
>>
>> Thank you
>>
>> -----Mensagem original-----
>> De: Eddie Drapkin [mailto:[email protected]]
>> Enviada: quarta-feira, 22 de Julho de 2009 13:11
>> Para: rszeus
>> Cc: [email protected]
>> Assunto: Re: [PHP] Replace in a string with regex
>>
>> On Wed, Jul 22, 2009 at 8:02 AM, rszeus <mailto:[email protected]>
>> <[email protected]> wrote:
>>
>>
>> Hello,
>>
>> I’m tryng to make some replacements on a string.
>>
>> Everything goês fine until the regular expression.
>>
>>
>>
>> $file = "screens/temp/7a45gfdi6icpan1jtb1j99o925_1_main.jpg";
>>
>> echo $a = str_replace(array(7a45gfdi6icpan1jtb1j99o925,
>> 'temp/',’_([0-9])’), array(“test”,"",””), $file)
>>
>>
>>
>> The idea is to remove /temp and the last _1 from the file name..but i’m only
>> getting this:
>>
>> screens/test_1_main.jpg
>>
>>
>>
>> I want it to be: screens/test_main.jpg
>>
>>
>>
>> Thank you
>>
>>
>>
>>
>>
>>
>> If you're trying to do a regular expression based search and replace,
>> you probably ought to use preg_replace instead of str_replace, as
>> str_replace doesn't parse regular expressions.
>>
>> Try this one out, I think I got what you wanted to do:
>>
>> <?php
>>
>> $file = "screens/temp/7a45gfdi6icpan1jtb1j99o925_1_main.jpg";
>>
>> echo preg_replace('#(screens/)temp/(.+?)_1(_main\.jpg)#', '$1$2$3', $file);
>>
>>
>>
>>
>> In the second parameter, $2 is the string you'd want to replace to
>> test so change '$1$2$3' to '$1test$3'.
>>
>> It seems like you're having trouble with regular expressions, may I
>> suggest you read up on them?
>>
>> http://www.regular-expressions.info/ is a pretty great free resource,
>> as ridiculous as the design is.
>>
>>
>>
>>
>>
>>
>>
>>
>> I tested this, with the double quotes and the curly braces around the
>> middle argument (to avoid PHP getting confused) and it didn't recognise
>> the matches by the numbered $1, $3, etc. I know I'm not the op who asked
>> the original question, but it might help him/her?
>>
>> Thanks
>> Ash
>> www.ashleysheridan.co.uk
>>
>>
>>
>>
>>
>> To avoid confusion, rather than something like
>> "$1{$id}$3"
>> Which looks really indecipherable, I'd definitely think something like
>> '$1' . $id . '$3'
>> is a lot easier to read and understand what's going on by immediately
>> looking at it.
>>
>> As an added bonus, it'll definitely work ;)
>>
>>
>>
>>
>
>
>
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