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:oorza2k5@xxxxxxxxx]
Enviada: quarta-feira, 22 de Julho de 2009 16:03
Para: ash@xxxxxxxxxxxxxxxxxxxx
Cc: Jim Lucas; rszeus; php-general@xxxxxxxxxxxxx
Assunto: Re: Replace in a string with regex
On Wed, Jul 22, 2009 at 11:01 AM, Ashley
Sheridan<ash@xxxxxxxxxxxxxxxxxxxx> 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:oorza2k5@xxxxxxxxx]
Enviada: quarta-feira, 22 de Julho de 2009 14:12
Para: rszeus
Cc: php-general@xxxxxxxxxxxxx
Assunto: Re: Replace in a string with regex
On Wed, Jul 22, 2009 at 9:07 AM, rszeus<rszeus@xxxxxxxxx> wrote:
Hi. It Works to remove the _1 but it doesn't replace '7a45gfdi6icpan1jtb1j99o925' for 'test'
Thank you
-----Mensagem original-----
De: Eddie Drapkin [mailto:oorza2k5@xxxxxxxxx]
Enviada: quarta-feira, 22 de Julho de 2009 13:11
Para: rszeus
Cc: php-general@xxxxxxxxxxxxx
Assunto: Re: Replace in a string with regex
On Wed, Jul 22, 2009 at 8:02 AM, rszeus<rszeus@xxxxxxxxx> 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 ;)
