Question

我目前有这段代码：

    if(event.getMessage().contains(plugin.getConfig().getString("PBSwears.words.word1")))
    {
        String message = event.getMessage().replaceAll(plugin.getConfig().getString("PBSwears.blocked.word1"), plugin.getConfig().getString("PBSwears.replace.word1"));
        event.setMessage(message);
    } 
    else if(event.getMessage().contains(plugin.getConfig().getString("PBSwears.words.word2")))
    {
        String message = event.getMessage().replaceAll(plugin.getConfig().getString("PBSwears.blocked.word2"), plugin.getConfig().getString("PBSwears.replace.word2"));
        event.setMessage(message);
    }

它将持续大约6倍......我正在尝试将其浓缩成类似的东西：

    String number = {find whitch number the message contains};
    if(event.getMessage().contains(plugin.getConfig().getString("PBSwears.words.word" + number)))
    {
        String message = event.getMessage().replaceAll(plugin.getConfig().getString("PBSwears.blocked.word" + number), plugin.getConfig().getString("PBSwears.replace.word" + number));
        event.setMessage(message);
    }

但我正在努力寻找包含这个词的号码（PBSwears.words.word1或PBSwears.words.word2）。我尝试过这样的事情：

    char number='0';
    if(event.getMessage().contains(plugin.getConfig().getString("PBSwears.words.word1")))
    {
        number='1';
    }
    else if(event.getMessage().contains(plugin.getConfig().getString("PBSwears.words.word2")))
    {
        number='2';
    }

但是再一次，如果有陈述，还有很多其他的。另外，我正在读取配置文件中的值，如下所示：

PBSwears:
  blocked:
    word1: *insert vulgar language*
    word2: *insert vulgar language*
  replace:
    word1: *insert funny replacement*
    word2: *insert funny replacement*

任何见解都会非常有用。

Answer 1

我建议加载一个地图，其中包含由被阻止的单词键入的替换单词。然后你可以利用这样的方法：

public Map<String, String> replacementWords = new HashMap<String, String>();
public String getWordReplacement(String word)
{
   if (replacementWords.containsKey(word) {
      return replacementWords.get(word);
   }

   return word;
}

您必须预先填充地图，类似于：

public void populateReplacementWordMap() {
   int numberOfWords = 6;
   for (int i = 1; i <= numberOfWords; i++) {
      String blockedWord = plugin.getConfig().getString("PBSwears.blocked.word" + i);
      String replacementWord = plugin.getConfig().getString("PBSwears.replace.word" + i);

      replacementWords.put(blockedWord, replacementWord);
   }
}

Answer 2

以下唯一的事情是必须定义MAX_COUNT。如果您知道自己要进行6次检查，请将其定义为6，依此类推。

public int getCorrectNumber() {
    for(int i = 1; i <= MAX_COUNT; i ++) {
        if(event.getMessage().contains(plugin.getConfig().getString("PBSwears.words.word" + i))) {
            return i;
        }
    }

    return -1;
}

然后

String number = "" + getCorrectNumber();

Answer 3

你为什么不迭代？

String[] badWords = {"word1", "word2", ...};
for (int i=0; i<badWords.length; i++){

  if(event.getMessage().contains(plugin.getConfig().getString("PBSwears.words." + badWords[i]))){
     String message = event.getMessage().replaceAll(plugin.getConfig().getString("PBSwears.blocked." + badWords[i]), plugin.getConfig().getString("PBSwears.replace." + badWords[i]));
     event.setMessage(message);
  }
}

试图压缩其他if语句

3 个答案: