捕获XSS(跨站点脚本)攻击的最佳正则表达式(在Java中)？

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

public class XSSRegex {
    public static void main(String[] args) {
        String input =<script>alert('XSS Attack!')</script>";
        String regex =<script[^>]*>([\\s\\S]*?)</script>";

        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(input);

        if (matcher.find()) {
            System.out.println("XSS Attack detected: " + matcher.group(0));
        } else {
            System.out.println("No XSS Attack detected");
        }
    }
}

public class XSSRegex {
    public static void main(String[] args) {
        String input =<script>alert('XSS Attack!')</script>";
        String regex =<script[^>]*>([\\s\\S]*?)</script>";

        String sanitizedInput = input.replaceAll(regex, "");

        System.out.println("Sanitized input: " + sanitizedInput);
    }
}