Capture everything with double quotes (optional) Regex javascript

Question

I have this regex right now:

file:\s?['"](.*?)['"]

For this string:

file: "word.124-ew/ef/?s=sf"

And the output is:

file: word.124-ew/ef/?s=sf

But I would also like to capture a group that hasn't double quotes (optional). For example, this:

file: word.124-ew/ef/?s=sf

No double quotes, but can also have them (optional).

I tried making this regex, but it hasn't the result desired:

file:\s?['"]?(.*)['"]?

And the output is:

word.124-ew/ef/?s=sf

With a final double quotes included, and its not what I expected (not to capture the double quotes).

Now with the last Regex I made, I have a problem when adding double quotes again. The output is:

file: word.124-ew/ef/?s=sf"

Tested with:

http://regexr.com/

RegExp: file:\s?['"]?(.*)['"]?

String1: file: "word.124-ew/ef/?s=sf"

String2: file: word.124-ew/ef/?s=sf

Fails on: file: word.124-ew/ef/?s=sf" (captures the final double quote and I don't want to capture the double quotes)

Any solutions? I will be glad if you guys can land me a hand! Thanks

Answer 1

You can use the following regex:

file:\s*['"]?([^'"]+)

See the regex demo (note that \n is added for the demo purposes only)

I used \s* to match 0 or more whitespace (if there is more than 1, it will match them all), added the ? quantifier to make ['"] optional, and turned the lazy dot matching subpattern into a negated character class ([^'"]+) matching 1 or more characters other than ' and ".