How to replace all double quotes without a backslash ahead

Question

Assume I have a string like below:

String param = "[\"\\n\",\"\\t\",\"'\",\"\\\"\",\"\\\\\"]"

The output of System.out.println is:

"\n","\t","'","\"","\\"

I would like to replace double quotes which doesn't have a backslash ahead, or, in another word, I would like to have the System.out.println output like below:

\n,\t,',\",\\

So I used this pattern:

System.out.println(param.replaceAll("\\\\{0}\"", ""));

But I got this:

\n,\t,',\,\\

As you can see, the double quote with a backslash ahead is also replaced. How can I prevent it from being replaced?

Edit: Sorry about the square brackets. You may ignore them cause they have nothing to do with this question

Answer 1

You can use the following regex to match and remove " that are string literal qualifiers:

(?s)(?<!\\)((?:\\{2})*)"([^"\\]*(?:\\.[^"\\]*)*)"

See the regex demo.

Details

• (?s) - DOTALL modifier (just in case the string literal can span across lines)
• (?<!\\) - no \ immediately to the left of the current location
• ((?:\\{2})*) - Group 1: any 0+ conseuctive occurrences of 2 backslashes
• " - a double quote (string literal start)
• ([^"\\]*(?:\\.[^"\\]*)*) - Group 2:
  • [^"\\]* - any 0+ chars other than \ and "
  • (?:\\.[^"\\]*)* - 0+ sequences of
    • \\. - a \ followed with any char
    • [^"\\]* - any 0+ chars other than \ and "
• " - a closing string literal double quote

See the Java demo:

String param = "[\"\\n\",\"\\t\",\"'\",\"\\\"\",\"\\\\\",\"\\\\\\\"\"]";
System.out.println(param);
// => ["\n","\t","'","\"","\\","\\\""]
String regex = "(?s)(?<!\\\\)((?:\\\\{2})*)\"([^\"\\\\]*(?:\\\\.[^\"\\\\]*)*)\"";
param = param.replaceAll(regex, "$1$2");
System.out.println(param);
// => [\n,\t,',\",\\,\\\"]