Regular expression to select all whitespace that isn't in quotes?

Question

I'm not very good at RegEx, can someone give me a regex (to use in Java) that will select all whitespace that isn't between two quotes? I am trying to remove all such whitespace from a string, so any solution to do so will work.

For example:

(this is a test "sentence for the regex")

should become

(thisisatest"sentence for the regex")

Answer 1

Here's a single regex-replace that works:

\s+(?=([^"]*"[^"]*")*[^"]*$) 

which will replace:

(this is a test "sentence for the regex" foo bar) 

with:

(thisisatest"sentence for the regex"foobar) 

Note that if the quotes can be escaped, the even more verbose regex will do the trick:

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

which replaces the input:

(this is a test "sentence \"for the regex" foo bar) 

with:

(thisisatest"sentence \"for the regex"foobar) 

(note that it also works with escaped backspaces: (thisisatest"sentence \\\"for the regex"foobar))

Needless to say (?), this really shouldn't be used to perform such a task: it makes ones eyes bleed, and it performs its task in quadratic time, while a simple linear solution exists.

EDIT

A quick demo:

String text = "(this is a test \"sentence \\\"for the regex\" foo bar)"; String regex = "\\s+(?=((\\\\[\\\\\"]|[^\\\\\"])*\"(\\\\[\\\\\"]|[^\\\\\"])*\")*(\\\\[\\\\\"]|[^\\\\\"])*$)"; System.out.println(text.replaceAll(regex, ""));  // output: (thisisatest"sentence \"for the regex"foobar)