字符串删失

时间:2013-10-28 22:25:23

标签: java word blacklist

我试图审查某些词语在游戏中被聊天。唯一的问题是玩家可以通过添加单词来取消我的审查。这是一个例子。

    //Check for rude words before sending to server
    List<String> tokens = new ArrayList<String>();
    tokens.add("bilbo");
    tokens.add("baggins");
    tokens.add("in");
    tokens.add("the");
    tokens.add("shire");

    String patternString = "\\b(" + StringUtils.join(tokens, "|") + ")\\b";
    Pattern pattern = Pattern.compile(patternString);
    Matcher findRudeWords = pattern.matcher(result.toLowerCase());

    while (findRudeWords.find()) {
        //Replace the bad word with astericks
        String asterisk = StringUtils.leftPad("", findRudeWords.group(1).length(), '*');
        result = result.replaceAll("(?i)" + findRudeWords.group(1), asterisk);
    }

常见问题是如果有人说bilbobaggins,中间没有空格,我的检查员可以很容易地避免。我怎么能做一个不仅仅检查单词的足够的审查?

1 个答案:

答案 0 :(得分:1)

取出两个字的边界。两个\ b's。我不想打扰StringUtils所需的额外库,所以我稍微修改了你的代码,但这是我测试过的:

import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.regex.*;

class Ideone {
    public static void main(String[] args) throws java.lang.Exception {
        // Check for rude words before sending to server
        String result = "heres bilbobaggins haha";
        String patternString = "(bilbo|baggins|in|the|shire)";
        Pattern pattern = Pattern.compile(patternString);
        Matcher findRudeWords = pattern.matcher(result.toLowerCase());

        while (findRudeWords.find()) {
            // Replace the bad word with asterisks
            result = result.replaceAll("(?i)" + findRudeWords.group(1), "*");
        }
        System.out.println("result=" + result);
    }

}

输出:

result=heres ** haha

你可以在这里玩:http://ideone.com/72SU7X