Java regular expression to match specific special characters

Question

I have input strings containing these special characters in any order: * . and \n

I want to make sure that there are no other characters in the string and \n should always be together and not something like *.*..\..n, so I want to match the string exactly in Java using a regular expression.

I tried using a regular expression to determine if an input string matches the pattern as below:

    String input = "*.*.*.\n..";
    System.out.println(input.matches("[\\\\.*\\n]"));

However, the output is false.

I tried using the double escape characters, in order to deal with Java's use of escape characters, but the result isn't as expected.

Answer 1

You just need to add the * quantifier to match more than one character. Also, there is no need to escape the literal dot:

String input = "*.*.*.\n..";
System.out.println(input.matches("[.*\\n]*"));

[.*\\n] matches a ., or a * or the literal newline character \n.