C# Regex pattern double quotes problem

Question

Hello all.

In the following code if the user string input is "a"b" or "a\"b" is match a"b string.

I don't expect that "a"b" is a valid regex pattern as the code return true

in the try catch block.

what is wrong with this code pattern.

Thank you.

  string? pattern = "";

  private string re_str = @"^\s*""(.*)""\s*.*$";

  //pattern_string is user string input

  private bool GetPattern(String pattern_string)

  {

    StringWriter swArrayToString = new StringWriter();

    string cleanPattern;

    int i;



    pattern = "";


        

    pattern_correct = true;

    string gstr = Regex.Match(pattern_string), re_str).Groups[1].Value;


   

    if(gstr != "") 

    {

     cleanPattern = Regex.Replace(RERichTextBoxLines[i], re_str, gstr);

     swArrayToString.Write(cleanPattern);

    }

    pattern = swArrayToString.ToString();

    if(String.IsNullOrEmpty(pattern))

    {

      return (false);

    }

    swArrayToString.Close();

    return (true);

  }

  private async void MatchButton_Clicked(object sender, EventArgs e)

  {

    GetPattern(user_string_input);

    try

    {

      Regex.Match("", pattern);
      return true;

    }

    catch

    {

      return false;

    }

  }

Answer 1

The issue with the code pattern lies in the regular expression itself and how it's being used.

Let's break down the regular expression @"^\s*""(.*)""\s*.*$":

• ^: Asserts the start of the string.
• \s*: Matches zero or more whitespace characters.
• "": Matches a double quotation mark.
• (.*): Captures any characters (except newline characters) between two double quotation marks.
• "": Matches a double quotation mark.
• \s*: Matches zero or more whitespace characters.
• .*: Matches any characters (except newline characters) zero or more times until the end of the string ($).

The intention of this regular expression seems to be to extract the content between double quotation marks. However, it's not correctly handling escaped double quotation marks within the string.

For example, if the input string is "a"b", the captured group will be a"b, which may not be the expected behavior. Similarly, for the input string "a\"b", the captured group will be a\b.

To fix this issue, you need to properly handle escaped double quotation marks in the regular expression. You can use the following regular expression:

@"^\s*""((?:[^""\\]|\\.)*)""\s*.*$"

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hth

Marcin