RegEx: Grabbing values between quotation marks

Question

I have a value like this "Foo Bar" "Another Value" something else

What RegEx expression will return the values enclosed in the quotation marks (e.g. Foo Bar and Another Value)?

Answer 1

In general, the following regular expression fragment is what you are looking for:

"(.*?)"

This uses the non-greedy *? operator to capture everything up to but not including the next double quote. Then, you use a language-specific mechanism to extract the matched text.

In Python, you could do:

>>> import re
>>> str = '"Foo Bar" "Another Value"'
>>> print re.findall(r'"(.*?)"', str)
['Foo Bar', 'Another Value']

Answer 2

I would go for:

"([^"]*)"

The [^"] is regex for any character except '"'
The reason I use this over the non greedy many operator is that I have to keep looking that up just to make sure I get it correct.

Answer 3

I've been using the following with great success:

([""'])(?:(?=(\\?))\2.)*?\1

It supports nested quotes as well.

Answer 4

echo 'junk "Foo Bar" not empty one "" this "but this" and this neither' | sed 's/[^\"]*\"\([^\"]*\)\"[^\"]*/>\1</g'

This will result in: >Foo Bar<><>but this<

Here I showed the result string between ><'s for clarity, also using the non-greedy version with this sed command we first throw out the junk before and after that ""'s and then replace this with the part between the ""'s and surround this by ><'s.

Answer 5

This version

• accounts for escaped quotes

• controls backtracking

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