Question

I want to create a RegEx that finds strings that begin and end in single or double quotes.

For example I can match such a case like this:

String: "Hello World"
RegEx: /[\"\'][^\"\']+[\"\']/

However, the problem occurs when quotes appear in the string itself like so:

String: "Hello" World"

We know the above expression will not work.

What I want to be able to do, it to have the escape within the string itself, since that will be functionality required anyway:

String: "Hello\" World"

Now I could come up with a long and complicated expression with various patterns in a group, one of them being:

RegEx: /[\"\'][^\"\']+(\\\"|\\\')+[^\"\']+[\"\']/

However that to me seems excessive, and I think there may be a shorter and more elegant solution.

Intended syntax:

run arg1 "arg1" "arg3 with \"" "\"arg4" "arg\"\"5"

As you can see, the quotes are really only used to make sure that string with spaces are counted as a single string. Do not worry about arg1, I should be able to match unquoted arguments.

I will make this easier, arguments can only be quoted using double-quotes. So i’ve taken single quotes out of the requirements of this question.

I have modified Rui Jarimba’s example:

/(?<=")(\\")*([^"]+((\\(\"))*[^"])+)((\\"")|")/

This now accounts pretty well for most cases, however there is one final case that can defeat this:

run -a "arg3 \" p2" "\"sa\"mple\"\\"

The second argument end with \\" which is a conventional way in this case to allow a backslash at the end of a nested string, unfortunately the regex thinks this is an escaped quote since the pattern \" still exists at the end of the pattern.

Answer 1

Try this regex:

['"]([^'"]+((\\(\"|'))*[^'"])+)['"]

Given the following string:

"Hello" World 'match 2' "wqwqwqwq wwqwqqwqw" no match here oopop "Hello \" World"

It will match

"Hello"
'match 2'
"wqwqwqwq wwqwqqwqw"
"Hello \" World"