RegEx: Grabbing values between quotation marks

Question

I have a value like this:

\"Foo Bar\" \"Another Value\" something else

What regex will return the

Answer 1

Unlike Adam's answer, I have a simple but worked one:

(["'])(?:\\\1|.)*?\1

And just add parenthesis if you want to get content in quotes like this:

(["'])((?:\\\1|.)*?)\1

Then $1 matches quote char and $2 matches content string.

Answer 2

I liked Axeman's more expansive version, but had some trouble with it (it didn't match for example

foo "string \\ string" bar

or

foo "string1"   bar   "string2"

correctly, so I tried to fix it:

# opening quote
(["'])
   (
     # repeat (non-greedy, so we don't span multiple strings)
     (?:
       # anything, except not the opening quote, and not 
       # a backslash, which are handled separately.
       (?!\1)[^\\]
       |
       # consume any double backslash (unnecessary?)
       (?:\\\\)*       
       |
       # Allow backslash to escape characters
       \\.
     )*?
   )
# same character as opening quote
\1

Answer 3

All the answer above are good.... except they DOES NOT support all the unicode characters! at ECMA Script (Javascript)

If you are a Node users, you might want the the modified version of accepted answer that support all unicode characters :

/(?<=((?<=[\s,.:;"']|^)["']))(?:(?=(\\?))\2.)*?(?=\1)/gmu

Try here.

Answer 4

I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:

(['"])(?:(?!\1|\\).|\\.)*\1

It does the trick and is still pretty simple and easy to maintain.

Demo (with some more test-cases; feel free to use it and expand on it).

---

_{PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:}

(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)

_{Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.}

_{Alternatively, modify the initial version by simply adding a group and extract the string form $2:}

(['"])((?:(?!\1|\\).|\\.)*)\1

_{PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.}

Answer 5

The pattern (["'])(?:(?=(\\?))\2.)*?\1 above does the job but I am concerned of its performances (it's not bad but could be better). Mine below it's ~20% faster.

The pattern "(.*?)" is just incomplete. My advice for everyone reading this is just DON'T USE IT!!!

For instance it cannot capture many strings (if needed I can provide an exhaustive test-case) like the one below:

$string = 'How are you? I\'m fine, thank you';

The rest of them are just as "good" as the one above.

If you really care both about performance and precision then start with the one below:

/(['"])((\\\1|.)*?)\1/gm

In my tests it covered every string I met but if you find something that doesn't work I would gladly update it for you.

Check my pattern in an online regex tester.

Answer 6

This version

• accounts for escaped quotes

• controls backtracking

  /(["'])((?:(?!\1)[^\\]|(?:\\\\)*\\[^\\])*)\1/