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Regex to filter BBCode lists



The Next CEO of Stack OverflowTempered Greedy Token - What is different about placing the dot before the negative lookaheadMatch all occurrences of a regexA comprehensive regex for phone number validationHow to negate specific word in regex?RegEx match open tags except XHTML self-contained tagsbbcode style tags with pregCheck whether a string matches a regex in JSparsing multiple lists in bbcode?Javascript regex on textareaRemove all instances of token from comma separated list with regexReplace text in HTML and BBCode sample










0















I am trying to improve legacy PHP code that deals with cleaning-up BBCode from a string and am currently facing trouble with lists.



The current solution for lists does the following:



...
$search[] = sprintf('~[%s](.*)[/%s]~smUi', 'list', 'list');
$search[] = sprintf('~[%s=(.*)](.*)[/%s]~smUi', 'list', 'list');
$search[] = sprintf('~[%s]~i', '*');
$replace[] = '$1';
$replace[] = '$2';
$replace[] = '';
...
return preg_replace($search, $replace, $string);


This works fine when the string is something like



[list]
[*]Item 1
[*]Item 2
[*]Item 3
[/list]


But it will also strip [*] if it's not inside lists and also fails with things like:



[list]
[*][list]
[*]Item 1.1
[*]Item 1.2
[*]Item 1.3
[/list]
[*]Item 2
[*]Item 3
[/list]


Is it even possible using RegExp only to strip the [list] or [list=1] + [/list] tags along with [*] if they are within lists?










share|improve this question




























    0















    I am trying to improve legacy PHP code that deals with cleaning-up BBCode from a string and am currently facing trouble with lists.



    The current solution for lists does the following:



    ...
    $search[] = sprintf('~[%s](.*)[/%s]~smUi', 'list', 'list');
    $search[] = sprintf('~[%s=(.*)](.*)[/%s]~smUi', 'list', 'list');
    $search[] = sprintf('~[%s]~i', '*');
    $replace[] = '$1';
    $replace[] = '$2';
    $replace[] = '';
    ...
    return preg_replace($search, $replace, $string);


    This works fine when the string is something like



    [list]
    [*]Item 1
    [*]Item 2
    [*]Item 3
    [/list]


    But it will also strip [*] if it's not inside lists and also fails with things like:



    [list]
    [*][list]
    [*]Item 1.1
    [*]Item 1.2
    [*]Item 1.3
    [/list]
    [*]Item 2
    [*]Item 3
    [/list]


    Is it even possible using RegExp only to strip the [list] or [list=1] + [/list] tags along with [*] if they are within lists?










    share|improve this question


























      0












      0








      0








      I am trying to improve legacy PHP code that deals with cleaning-up BBCode from a string and am currently facing trouble with lists.



      The current solution for lists does the following:



      ...
      $search[] = sprintf('~[%s](.*)[/%s]~smUi', 'list', 'list');
      $search[] = sprintf('~[%s=(.*)](.*)[/%s]~smUi', 'list', 'list');
      $search[] = sprintf('~[%s]~i', '*');
      $replace[] = '$1';
      $replace[] = '$2';
      $replace[] = '';
      ...
      return preg_replace($search, $replace, $string);


      This works fine when the string is something like



      [list]
      [*]Item 1
      [*]Item 2
      [*]Item 3
      [/list]


      But it will also strip [*] if it's not inside lists and also fails with things like:



      [list]
      [*][list]
      [*]Item 1.1
      [*]Item 1.2
      [*]Item 1.3
      [/list]
      [*]Item 2
      [*]Item 3
      [/list]


      Is it even possible using RegExp only to strip the [list] or [list=1] + [/list] tags along with [*] if they are within lists?










      share|improve this question
















      I am trying to improve legacy PHP code that deals with cleaning-up BBCode from a string and am currently facing trouble with lists.



      The current solution for lists does the following:



      ...
      $search[] = sprintf('~[%s](.*)[/%s]~smUi', 'list', 'list');
      $search[] = sprintf('~[%s=(.*)](.*)[/%s]~smUi', 'list', 'list');
      $search[] = sprintf('~[%s]~i', '*');
      $replace[] = '$1';
      $replace[] = '$2';
      $replace[] = '';
      ...
      return preg_replace($search, $replace, $string);


      This works fine when the string is something like



      [list]
      [*]Item 1
      [*]Item 2
      [*]Item 3
      [/list]


      But it will also strip [*] if it's not inside lists and also fails with things like:



      [list]
      [*][list]
      [*]Item 1.1
      [*]Item 1.2
      [*]Item 1.3
      [/list]
      [*]Item 2
      [*]Item 3
      [/list]


      Is it even possible using RegExp only to strip the [list] or [list=1] + [/list] tags along with [*] if they are within lists?







      php regex






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 8 at 16:33









      benvc

      6,4531827




      6,4531827










      asked Mar 8 at 16:08









      smaressmares

      406518




      406518






















          1 Answer
          1






          active

          oldest

          votes


















          0














          You may use



          $search[] = sprintf('~[(%s)(?:=[^]]*)?]((?:(?![1b).)*?)[/1]s*~si', 'list');
          $search[] = sprintf('~[%s]~i', '\*');
          $replace[] = '$2';
          $replace[] = '';
          $count = 0;
          do
          $string = preg_replace($search, $replace, $string, -1, $count);

          while ($count > 0);
          return $string;


          See the PHP demo.



          I merged the first two regexps since they basically match the same (the =.*? part inside the open tag is just optional, I suggest using (?:=[^]]*)? to match = and then 0+ chars other than ] 1 or 0 times.



          The ((?:(?![1b).)*?) pattern is a tempered greedy token that ensures that the innermost list tag is matched, 1 here matches the tag name that is captured with (%s).



          The $count variable will hold the number of replacements done with preg_replace, and if nothing is replaced, the while block exits.






          share|improve this answer























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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            You may use



            $search[] = sprintf('~[(%s)(?:=[^]]*)?]((?:(?![1b).)*?)[/1]s*~si', 'list');
            $search[] = sprintf('~[%s]~i', '\*');
            $replace[] = '$2';
            $replace[] = '';
            $count = 0;
            do
            $string = preg_replace($search, $replace, $string, -1, $count);

            while ($count > 0);
            return $string;


            See the PHP demo.



            I merged the first two regexps since they basically match the same (the =.*? part inside the open tag is just optional, I suggest using (?:=[^]]*)? to match = and then 0+ chars other than ] 1 or 0 times.



            The ((?:(?![1b).)*?) pattern is a tempered greedy token that ensures that the innermost list tag is matched, 1 here matches the tag name that is captured with (%s).



            The $count variable will hold the number of replacements done with preg_replace, and if nothing is replaced, the while block exits.






            share|improve this answer



























              0














              You may use



              $search[] = sprintf('~[(%s)(?:=[^]]*)?]((?:(?![1b).)*?)[/1]s*~si', 'list');
              $search[] = sprintf('~[%s]~i', '\*');
              $replace[] = '$2';
              $replace[] = '';
              $count = 0;
              do
              $string = preg_replace($search, $replace, $string, -1, $count);

              while ($count > 0);
              return $string;


              See the PHP demo.



              I merged the first two regexps since they basically match the same (the =.*? part inside the open tag is just optional, I suggest using (?:=[^]]*)? to match = and then 0+ chars other than ] 1 or 0 times.



              The ((?:(?![1b).)*?) pattern is a tempered greedy token that ensures that the innermost list tag is matched, 1 here matches the tag name that is captured with (%s).



              The $count variable will hold the number of replacements done with preg_replace, and if nothing is replaced, the while block exits.






              share|improve this answer

























                0












                0








                0







                You may use



                $search[] = sprintf('~[(%s)(?:=[^]]*)?]((?:(?![1b).)*?)[/1]s*~si', 'list');
                $search[] = sprintf('~[%s]~i', '\*');
                $replace[] = '$2';
                $replace[] = '';
                $count = 0;
                do
                $string = preg_replace($search, $replace, $string, -1, $count);

                while ($count > 0);
                return $string;


                See the PHP demo.



                I merged the first two regexps since they basically match the same (the =.*? part inside the open tag is just optional, I suggest using (?:=[^]]*)? to match = and then 0+ chars other than ] 1 or 0 times.



                The ((?:(?![1b).)*?) pattern is a tempered greedy token that ensures that the innermost list tag is matched, 1 here matches the tag name that is captured with (%s).



                The $count variable will hold the number of replacements done with preg_replace, and if nothing is replaced, the while block exits.






                share|improve this answer













                You may use



                $search[] = sprintf('~[(%s)(?:=[^]]*)?]((?:(?![1b).)*?)[/1]s*~si', 'list');
                $search[] = sprintf('~[%s]~i', '\*');
                $replace[] = '$2';
                $replace[] = '';
                $count = 0;
                do
                $string = preg_replace($search, $replace, $string, -1, $count);

                while ($count > 0);
                return $string;


                See the PHP demo.



                I merged the first two regexps since they basically match the same (the =.*? part inside the open tag is just optional, I suggest using (?:=[^]]*)? to match = and then 0+ chars other than ] 1 or 0 times.



                The ((?:(?![1b).)*?) pattern is a tempered greedy token that ensures that the innermost list tag is matched, 1 here matches the tag name that is captured with (%s).



                The $count variable will hold the number of replacements done with preg_replace, and if nothing is replaced, the while block exits.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 8 at 17:13









                Wiktor StribiżewWiktor Stribiżew

                327k16147226




                327k16147226





























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