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

java regex escaping
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