Question

177

Regex to match all instances not inside quotes

rated 0 times [ 180] [ 3] / answers: 1 / hits: 54512 / 13 Years ago, fri, june 24, 2011, 12:00:00

From this q/a, I deduced that matching all instances of a given regex not inside quotes, is impossible. That is, it can't match escaped quotes (ex: "this whole "match" should be taken"). If there is a way to do it that I don't know about, that would solve my problem.

If not, however, I'd like to know if there is any efficient alternative that could be used in JavaScript. I've thought about it a bit, but can't come with any elegant solutions that would work in most, if not all, cases.

Specifically, I just need the alternative to work with .split() and .replace() methods, but if it could be more generalized, that would be the best.

For Example:

An input string of:
+bar+baz"not+or"+or+"this+"foo+bar+

replacing + with #, not inside quotes, would return:
#bar#baz"not+or"+or+"this+"foo#bar#

Answers

Only authorized users can answer the question. Please sign in first, or register a free account.

elaine

Add To Favorites

Follow

Total Points: 628

Total Questions: 111

Total Answers: 104

Location: Marshall Islands

Member since Tue, Sep 21, 2021

3 Years ago

answered 13 Years ago shylaelisan · Accepted Answer

Actually, you can match all instances of a regex not inside quotes for any string, where each opening quote is closed again. Say, as in you example above, you want to match +.

The key observation here is, that a word is outside quotes if there are an even number of quotes following it. This can be modeled as a look-ahead assertion:

+(?=([^]*[^]*)*[^]*$)

Now, you'd like to not count escaped quotes. This gets a little more complicated. Instead of [^]* , which advanced to the next quote, you need to consider backslashes as well and use [^\]*. After you arrive at either a backslash or a quote, you need to ignore the next character if you encounter a backslash, or else advance to the next unescaped quote. That looks like (\.|([^\]*\.)*[^\]*). Combined, you arrive at

+(?=([^\]*(\.|([^\]*\.)*[^\]*))*[^]*$)

I admit it is a little cryptic. =)