Get code outside quotes with regexp in javascript

Question

Is there a way to get a piece of code that is not between quotes (single or double) in javascript with regular expressions?

if i have this string:

'this is a test "this shouldn't be taken"'

the result should be:

'this is a test'

Answer 1

myString.replace(/".*?"/g, '')

will remove any string between double-quotes from myString. It doesn't handle escaped double-quotes, though.

Answer 2

You could remove the quoted part of your string with the javascript replace function:

str = 'this is a test "this shouldn\'t be taken"';
str_without_quotes = str.replace(/(['"]).*?\1/g, "") // => 'this is a test '

Answer 3

This should remove anything between single or double quotes, it works with multi-line strings (strings that contain \n or \r) and it should also handle escaped quotes:

var removeQuotes = /(['"])(?:\\?[\s\S])*?\1/g;

var test = 'this is a test "this shouldn\'t be taken"';
test.replace(removeQuotes, ""); // 'this is a test '

test = 'this is a test "this sho\\"uldn\'t be taken"';
test.replace(removeQuotes, ""); // 'this is a test '

Answer 4

thanks Prestaul your solution works well