Re: Re[2]: [PHP] preg_replace: avoiding double replacements

On 18 May 2010 09:04, Andre Polykanine <andre@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 <b<newtag>> with the <<b<newtag>>tag> 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>