Java 试图压缩else if语句

我目前有以下代码：

    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次…我试图把它浓缩成这样的东西：

    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);
    } 

但我正在努力寻找单词包含在哪个数字中（PBSwears.words.word1或PBSwears.words.word2）。我试过这样的方法：

    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';
    }

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

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

   return word;
}

但是，还有大量的else-if语句。 另外，我正在从一个配置文件中读取如下所示的值：

    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';
    }

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

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

   return word;
}

---

任何洞察都会非常有用。

以下内容中唯一的一点是必须定义

MAX\u COUNT

。如果您知道要检查6次，那么将其定义为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;
}

你为什么不迭代呢

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);
  }
}

String[]badWords={“word1”、“word2”、…}；
对于（int i=0；i，我建议加载一个映射，其中包含由阻塞字键入的替换字。然后，您可以使用如下方法：

    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';
    }

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

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

   return word;
}

publicmap replacementWords=newhashmap（）；
公共字符串getWordReplacement（字符串字）
{
if（替换词.containsKey（词）{
返回replacementWords.get（word）；
}
返回词；
}

您必须预先填充地图，如下所示：

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);
   }
}

public void populateReplacementWordMap（）{
int numberOfWords=6；
对于（int i=1；i听起来，这其中的大部分可以重构为一个helper方法，其变化通过方法参数表示。除非MAX_COUNT=1@Salauyou哎哟，那是个打字错误。刚刚修好。谢谢。这个工作做得很好。真不敢相信我想不出像这样的东西。顺便说一句，看看如何使用
 HashMap
s，如下所述，在进一步的重构中。当然，它会让您更改API，但在性能方面要好得多。