Find and print illegal character in a string using regexp

Question

I have the following simple regexp that checks for illegal characters in a string:
String regexp = ".*([\\[\\];]+|@).*"; // illegal: [ ] ; @
String input = "Testing [ 123";
System.out.println(Pattern.matches(regexp, input));How can I find and print the character that is invalid??
I've tried using the Matcher class togheter with Pattern but cant get it to work. :(
Like this:
Pattern pattern = Pattern.compile(regexp);
Matcher matcher = pattern.matcher(input);
matcher.lookingAt();
int matchedChar = matcher.end();
if (matchedChar < input.length()) {
    String illegalCharFound = String.valueOf(input.charAt(matcher.end()));
    System.out.println(illegalCharFound);
}What am I doing wrong?

Answer 1

1. You call lookingAt(), but you don't check its return value, so you don't know if the regex actually matched.
2. matcher.end() returns the index of the character following whatever was matched, assuming there was a match (if there wasn't, it will throw an exception). So either it will point to a legal character, or you'll get a StringIndexOutOfBoundsException because an illegal character was found at the end of the input. The start() method would be a better choice, but...
3. Your regex can match multiple consecutive square brackets or semicolons (and why not put the at-sign in the character class, too?), but you're acting like it can only match one character. Even if there is only one character, group(1) is an easier way to extract it. Also, if there are more than one (non-consecutive) illegal characters, your regex will only find the last one. That's because the first dot-star initially gobbles up the whole input, then backtracks only as far as it has to to satisfy the rest of the regex. If your goal is to provide feedback to whoever supplied the input, it's going to be pretty confusing feedback. You should probably use the find() method in a loop to pick out all the illegal characters so you can report them properly.