Re[2]: [PHP] preg_replace: avoiding double replacements
Hello Jim, That might work for that particular example, but I have utf-8 strings containing different characters of different alphabets, so neither str_replace nor strtr work... Thanks! -- With best regards from Ukraine, Andre Skype: Francophile; WlmMSN: arthaelon @ yandex.ru; Jabber: arthaelon @ jabber.org Yahoo! messenger: andre.polykanine; ICQ: 191749952 Twitter: m_elensule - Original message - From: Jim Lucas li...@cmsws.com To: Andre Polykanine an...@oire.org Date: Tuesday, May 18, 2010, 3:33:09 AM Subject: [PHP] preg_replace: avoiding double replacements Andre Polykanine wrote: Hello everyone, Sorry for bothering you again. Today I met a problem exactly described by a developer in users' notes that follow the preg_replace description in the manual: info at gratisrijden dot nl 02-Oct-2009 02:48 if you are using the preg_replace with arrays, the replacements will apply as subject for the patterns later in the array. This means replaced values can be replaced again. Example: ?php $text = 'We want to replace BOLD with the boldtag and OLDTAG with the newtag'; $patterns = array( '/BOLD/i', '/OLDTAG/i'); $replacements = array( 'boldtag', 'newtag'); echo preg_replace ($patterns, $replacements, $text); ? Output: We want to replace bnewtag with the bnewtagtag and newtag with the newtag Look what happend with BOLD. Is there any solution to this besides any two-step sophisticated trick like case changing? Thanks! -- With best regards from Ukraine, Andre Http://oire.org/ - The Fantasy blogs of Oire Skype: Francophile; WlmMSN: arthaelon @ yandex.ru; Jabber: arthaelon @ jabber.org Yahoo! messenger: andre.polykanine; ICQ: 191749952 Twitter: http://twitter.com/m_elensule Well, for the example you gave, why use regex? Check this out as an example. plaintext?php $text = 'We want to replace BOLD with the boldtag and OLDTAG with the newtag'; $regex = array( '/BOLD/i', '/OLDTAG/i', ); $oldtags = array( 'BOLD', 'OLDTAG', ); $replacements = array( 'boldtag', 'newtag', ); # Original String echo $text.\n; # After regex is applied echo preg_replace($regex, $replacements, $text).\n; # After plain tag replacement happens echo str_replace($oldtags, $replacements, $text).\n; ? See if that works for you. -- Jim Lucas Some men are born to greatness, some achieve greatness, and some have greatness thrust upon them. Twelfth Night, Act II, Scene V by William Shakespeare -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: Re[2]: [PHP] preg_replace: avoiding double replacements
On 18 May 2010 09:04, Andre Polykanine an...@oire.org wrote: [snip] Andre Polykanine wrote: Hello everyone, Sorry for bothering you again. Today I met a problem exactly described by a developer in users' notes that follow the preg_replace description in the manual: info at gratisrijden dot nl 02-Oct-2009 02:48 if you are using the preg_replace with arrays, the replacements will apply as subject for the patterns later in the array. This means replaced values can be replaced again. Example: ?php $text = 'We want to replace BOLD with the boldtag and OLDTAG with the newtag'; $patterns = array( '/BOLD/i', '/OLDTAG/i'); $replacements = array( 'boldtag', 'newtag'); echo preg_replace ($patterns, $replacements, $text); ? Output: We want to replace bnewtag with the bnewtagtag and newtag with the newtag Look what happend with BOLD. Is there any solution to this besides any two-step sophisticated trick like case changing? Thanks! Use better regexes: either match for word endings or use a delimiter in your markers (i.e. ###BOLD### instead of BOLD). Regards Peter -- hype WWW: http://plphp.dk / http://plind.dk LinkedIn: http://www.linkedin.com/in/plind Flickr: http://www.flickr.com/photos/fake51 BeWelcome: Fake51 Couchsurfing: Fake51 /hype -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re[2]: [PHP] preg_replace();
I have tasted the code and it worked fine (if I got you right):
$old_string=lazy \|\ dog;
$new_string=str_replace('|', '_', $old_string);
print $new_string;
I got lazy_dog
Ed
Friday, February 2, 2007, 10:01:14 PM, you wrote:
Thanks,
but I think that I must use preg_replace because the condition is: replace
the chars (pipe or space) when they are between
ie : src=file:///h|/hjcjdgh dlkgj/dgjk.jpg to
src=file:///h_/hjcjdgh_dlkgj/dgjk.jpg
Seb
- Original Message -
From: [EMAIL PROTECTED]
To: Sébastien WENSKE [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Friday, February 02, 2007 8:38 PM
Subject: Re: [PHP] preg_replace();
I am not a very experienced programmer, but I think that str_replace
can be used in this case:
$new_string=str_replace('|', '_', $old_string)
then use the same function to replace spaces.
Ed
Friday, February 2, 2007, 9:30:37 PM, you wrote:
Hi all,
I want replace the | (pipe) and the (space) chars where are
between (double-quotes) by an underscore _ with the
preg_replace(); funtction.
Can someone help me to find the correct regex.
Thanks in advance
Seb
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