Remove all punctuation from the end of a string

Question

Examples:

// A B C.       -> A B C
// !A B C!      -> !A B C
// A? B?? C???  -> A? B?? C

Here's what I have so far:

while (endsWithRegex(word, "\\p{P}")) {
    word = word.substring(0, word.length() - 1);
}

public static boolean endsWithRegex(String word, String regex) {
    return word != null && !word.isEmpty() && 
        word.substring(word.length() - 1).replaceAll(regex, "").isEmpty();
}

This current solution works, but since it's already calling String.replaceAll within endsWithRegex, we should be able to do something like this:

word = word.replaceAll(/* regex */, "");

Any advice?

Answer 1

I suggest using

\s*\p{Punct}+\s*$

It will match optional whitespace and punctuation at the end of the string.

If you do not care about the whitespace, just use \p{Punct}+$.

Do not forget that in Java strings, backslashes should be doubled to denote literal backslashes (that must be used as regex escape symbols).

Java demo

String word = "!Words word! ";
word = word.replaceAll("\\s*\\p{Punct}+\\s*$", "");
System.out.println(word); // => !Words word