正则表达式 - 列出与模式不匹配的字符

Question

I have below regex which says if the given input matches the pattern or not

String input="The input here @$%/";
String pattern="[a-zA-Z0-9,.\\s]*";
if (!input.matches(pattern)) {
System.out.println("not matched");
}else{
  System.out.println("matched");
}

Can i know how it can be enhanced to list the characters in the input, which are not matching the pattern. Eg here @$%/

Answer 1

As anubhava has already mentioned in the comment , just use input.replaceAll(pattern, "") .

Demo:

class Main {
    public static void main(String[] args) {
        String input = "The input here @$%/";
        String pattern = "[a-zA-Z0-9,.\\s]*";
        String nonMatching = input.replaceAll(pattern, "");
        System.out.println(nonMatching);
    }
}

Output:

@$%/

Answer 2

Use

[^a-zA-Z0-9,.\s]

See regex proof .

EXPLANATION

 [^a-zA-Z0-9,.\s]         any character except: 'a' to 'z', 'A' to
                           'Z', '0' to '9', ',', '.', whitespace (\n,
                           \r, \t, \f, and " ")

Java code snippet :

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Example {
    public static void main(String[] args) {
        final String regex = "[^a-zA-Z0-9,.\\s]";
        final String string = "The input here @$%/";
        
        final Pattern pattern = Pattern.compile(regex);
        final Matcher matcher = pattern.matcher(string);
        
        while (matcher.find()) {
            System.out.println(matcher.group(0));
        }
    }
}