Question

I’m using php’s preg_replace function, and I have the following regex:

(?:[^>(),]+) 

to match any characters but >(),. The problem is that I want to make sure that there is at least one letter in it (\w) and the match is not empty, how can I do that?

Is there a way to say what i DO WANT to match in the [^>(),]+ part?

Answer 1

You can add a lookahead assertion:

(?:(?=.*\p{L})[^>(),]+)

This makes sure that there will be at least one letter (\p{L}; \w also matches digits and underscores) somewhere in the string.

You don’t really need the (?:...) non-capturing parentheses, though:

(?=.*\p{L})[^>(),]+

works just as well. Also, to ensure that we always match the entire string, it might be a good idea to surround the regex with anchors:

^(?=.*\p{L})[^>(),]+$

EDIT:

For the added requirement of not including surrounding whitespace in the match, things get a little more complicated. Try

^(?=.*\p{L})(\s*)((?:(?!\s*$)[^>(),])+)(\s*)$

In PHP, for example to replace all those strings we found with REPLACEMENT, leaving leading and trailing whitespace alone, this could look like this:

$result = preg_replace(
    '/^          # Start of string
    (?=.*\p{L})  # Assert that there is at least one letter
    (\s*)        # Match and capture optional leading whitespace  (--> \1)
    (            # Match and capture...                           (--> \2)
     (?:         # ...at least one character of the following:
      (?!\s*$)   # (unless it is part of trailing whitespace)
      [^>(),]    # any character except >(),
     )+          # End of repeating group
    )            # End of capturing group
    (\s*)        # Match and capture optional trailing whitespace (--> \3)
    $            # End of string
    /xu', 
    '\1REPLACEMENT\3', $subject);