Escaping entire String in a Regular Expression

Question

How does one escape an entire String in a Pattern? I thought that prefixing it with "\\Q" and postfixing with "\\E" would do the trick but this is functioning strangely. It also ignores the possibility of a "\\Q" and/or "\\E" within the String. I guess one could escape every meta-character but this seems like overkill. Is there a utility method somewhere? Why is it that:
System.out.println("xx\\xx".replaceAll("\\Qx\\x\\E", "a"));yields xax but
System.out.println("xx\\xx".replaceAll("\\Q\\\\E", "a"));outputs xx\xx?

Answer 1

It looks like you've uncovered a bug. Specifically, when the Pattern parser sees the backslash that you're trying to match, it looks at the next character to see if it's an 'E'. It isn't, so the parser consumes both characters, then starts looking for the \E sequence again. The next character, of course, is the 'E', but the parser just sees it as a literal 'E' now. End result: instead of a Pattern for a single backslash, you get a Pattern for a backslash, followed by a backslash, followed by an 'E', as demonstrated here:    System.out.println("xx\\\\Exx".replaceAll("\\Q\\\\E", "a")); prints xxaxx.
Since you're escaping the whole regex, you can work around the bug by leaving off the \E:    System.out.println("xx\\xx".replaceAll("\\Q\\", "a")); prints xxaxx.
Or you can do what I do and use this method to escape strings instead of \B ... \E:  public static String quotemeta(String str)
    if (str.length() == 0)
      return "";
    StringBuffer buf = new StringBuffer();
    for (int i = 0; i < str.length(); i++)
      char c = str.charAt(i);
      if ("\\[](){}.*+?$^|".indexOf(c) != -1)
        buf.append('\\');
      buf.append(c);
    return buf.toString();
  }(This is why I've never run afoul of this bug, though I use regexes a lot.)
I'll submit this to BugParade if you like. While I'm at it, I can submit an RFE to have them add a quotemeta method to Pattern.