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Is there any method in Java or any open source library for escaping (not quoting) a special character (meta-character), in order to use it as a regular expression? This would be very handy in dynamically building a regular expression, without having to manually escape each individual character. For example, consider a simple regex like \d+\.\d+ that matches numbers with a decimal point like 1.2 , as well as the following code: String digit = "d"; String point = "."; String regex1 = "\\d+\\.\\d+"; String regex2 = Pattern.quote(digit + "+" + point + digit + "+"); Pattern numbers1 = Pattern.compile(regex1); Pattern numbers2 = Pattern.compile(regex2); System.out.println("Regex 1: " + regex1); if (numbers1.matcher("1.2").matches()) { System.out.println("\tMatch"); } else { System.out.println("\tNo match"); } System.out.println("Regex 2: " + regex2); if (numbers2.matcher("1.2").matches()) { System.out.println("\tMatch"); } else { System.out.println("\tNo match"); } Not surprisingly, the output produced by the above code is: Regex 1: \d+\.\d+ Match Regex 2: \Qd+.d+\E No match That is, regex1 matches 1.2 but regex2 (which is "dynamically" built) does not (instead, it matches the literal string d+.d+ ). So, is there a method that would automatically escape each regex meta-character? If there were, let's say, a static escape() method in java.util.regex.Pattern , the output of Pattern.escape('.') would be the string "\." , but Pattern.escape(',') should just produce "," , since it is not a meta-character. Similarly, Pattern.escape('d') could produce "\d" , since 'd' is used to denote digits (althoug

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