Find an replace Guid to binary equivalent with regular expression

Question

I have a string containing a regular SQL-Statement with Guids. The statement should be converted so I can run it against a local SQLite database which cannot handle Guids as described by .NET-Framework. Therefore I have to convert them to their binary representation, which means that the Guid (in SQL) '00000000-0000-0000-0000-000000000000' will become X'00000000000000000000000000000000' for SQLite or 'FE334797-0A46-468D-91F2-0005F1EC67EC' will become X'974733fe460a8d4691f20005f1ec67ec'.

The method to convert a single Guid is as follows:

private static string GetBinaryGuid(Guid guidToConvert)
{
    var guidBytes = guidToConvert.ToByteArray();
    var guidBinary = new StringBuilder();
    foreach (var guidByte in guidBytes)
    {
        guidBinary.AppendFormat(@"{0}", guidByte.ToString("x2"));
    }
    return guidBinary.ToString();
}

The method to find the real Guid(s) in the query-string is:

resultString = Regex.Replace(subjectString, @"\b[A-F0-9]{8}(?:-[A-F0-9]{4}){3}-[A-F0-9]{12}\b", "'$0'", RegexOptions.IgnoreCase);

My question is how to replace the "real" Guid(s) in the string with their respective binary equivalent?

Edit: Just to clarify. I want to fetch all Guid in the string, pass the found Guids to the method mentioned above and replace it within the string. The result should be the SQL-Query with the binary Guid(s) if found in the string.

Edit2:

SELECT * FROM table WHERE Id = 'FE334797-0A46-468D-91F2-0005F1EC67EC'

should become

SELECT * FROM table WHERE Id = X'974733fe460a8d4691f20005f1ec67ec'

Edit3:

@aelor got me the right direction. For my specific case the solution can be found here.

Answer 1

i know this is very big but thats what I could come up with:

\b([A-F0-9]{2})([A-F0-9]{2})([A-F0-9]{2})([A-F0-9]{2})-([A-F0-9]{2})([A-F0-9]{2})-([A-F0-9]{2})([A-F0-9]{2})-([A-F0-9]{2})([A-F0-9]{2})-([A-F0-9]{12})\b

use this and you will get your result in matched groups.

The replacement string will look like this :

X'\4\3\2\1\6\5\8\7\9\10\11'

use a \L for making it lowercase.\

Demo here

if you are having trouble like this :

'X'974733FE460A91F20005F1EC67EC''

you can easily remove the leading and trailing ' by using a function

public class Main {   
  /**
   * Remove the leading and trailing quotes from <code>str</code>.
   * E.g. if str is '"one two"', then 'one two' is returned.
   *
   *
   * @return The string without the leading and trailing quotes.
   */
  static String stripLeadingAndTrailingQuotes(String str)
  {
      if (str.startsWith("\'"))
      {
          str = str.substring(1, str.length());
      }
      if (str.endsWith("\'"))
      {
          str = str.substring(0, str.length() - 1);
      }
      return str;
  }

}