Javascript Regex for all words not between certain characters

Question

I\'m trying to return a count of all words NOT between square brackets. So given ..

[don\'t match these words] but do match these

I get a c

Answer 1

Ok, I think this should work:

\[[^\]]+\](?:^|\s)([\w']+)(?!\])\b|(?:^|\s)([\w']+)(?!\])\b

You can test it here:
http://regexpal.com/

If you need an alternative with text in square brackets coming after the main text, it could be added as a second alternative and the current second one would become third.
It's a bit complicated but I can't think of a better solution right now.

If you need to do something with the actual matches you will find them in the capturing groups.

UPDATE:

Explanation: So, we've got two options here:

1. \[[^\]]+\](?:^|\s)([\w']+)(?!\])\b

This is saying:

• \[[^\]]+\] - match everything in square brackets (don't capture)
• (?:^|\s) - followed by line start or a space - when I look at it now take the caret out as it doesn't make sense so this will become just \s
• ([\w']+) - match all following word characters as long as (?!\])the next character is not the closing bracket - well this is probably also unnecessary now, so let's try and remove the lookahead
• \b - and match word boundary

2 (?:^|\s)([\w']+)(?!\])\b

If you cannot find the option 1 - do just the word matching, without looking for square brackets as we ensured with the first part that they are not here.

Ok, so I removed all the things that we don't need (they stayed there because I tried quite a few options before it worked:-) and the revised regex is the one below:

\[[^\]]+\]\s([\w']+)(?!\])\b|(?:^|\s)([\w']+)\b