Regex replace word when not enclosed in brackets

Question

I\'m trying to create a regular expression where it replaces words which are not enclosed by brackets.

Here is what I currently have:

$this->parse         


        

Answer 1

After another morning of playing with the regex I came up with a quite dirty solution which isn't flexible at all, but works for my use case.

$this->parsed = preg_replace('/\b(?!\[(|((\w+)(\s|\.))|((\w+)(\s|\.)(\w+)(\s|\.))))('.preg_quote($word).')(?!(((\s|\.)(\w+))|((\s|\.)(\w+)(\s|\.)(\w+))|)\[)\b/s','[$10['.implode(",",array_unique($types)).']]',$this->parsed);

What it basically does is check for brackets with no words, 1 word or 2 words in front or behind it in combination with the specified keyword.

Still, it would be great to hear if anyone has a better solution.

Answer 2

You may match any substring inside parentheses with \[[^][]*] pattern, and then use (*SKIP)(*FAIL) PCRE verbs to drop the match, and only match your pattern in any other context:

\[[^][]*](*SKIP)(*FAIL)|your_pattern_here

See the regex demo. To skip matches inside paired nested square brackets, use a recusrsion-based regex with a subroutine (note it will have to use a capturing group):

(?<skip>\[(?:[^][]++|(?&skip))*])(*SKIP)(*FAIL)|your_pattern_here

See a regex demo

Also, since you are building the pattern dynamically, you need to preg_quote the $word along with the delimiter symbol (here, /).

Your solution is

$this->parsed = preg_replace(
    '/\[[^][]*\[[^][]*]](*SKIP)(*FAIL)|\b(?:' . preg_quote($word, '/') . ')\b/', 
    '[$0[' . implode(",", array_unique($types)) . ']]',
    $this->parsed);

The \[[^][]*\[[^][]*]] regex will match all those occurrences that have been wrapped with your replacement pattern:

• \[ - a [
• [^][]* - 0+ chars other than [ and ]
• \[ - a [ char
• [^][]* - 0+ chars other than [ and ]
• ]] - a ]] substring.