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PHP ereg() Posted by jarudin (jarudin), 25 April 2004 I made a script to convert a BBcode like system into normal HTML (see code below)The only problem is that repetion of the same code (like in the string I inputted in the script the script will take the first b and very last /b and convert those to <b> and </b>. Which means the 'count' is greedy. I found a way around it, but cant seem to implemtent it. http://www.wellho.net/regex/php.html 'Add a ? after any count to turn it sparse (match as few as possible) rather than have it default to greedy' I dont understand where to put the '?' I'va already tried working around the '?' but that only made it worse. Code:
Thanks, --Jarudin Posted by admin (Graham Ellis), 26 April 2004 The short answer is that the ? goes straight after the regular count operator. Thus:+ one or more (as many as possible) +? one or moer (as few as possible) {2,6} from 2 to 6, as many as possible {2,6}? from 2 to 6, as few as possible and so on. You have things line +?, *?, ??, {3,}? and {4,6}? ..... This page is a thread posted to the opentalk forum
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