Question

下面我有以下课程。如何获取wordValidation（）方法以提供wordContent，以仅为找到的单词及其索引提供输出。对于使用我的测试程序下面的示例输出，如果我有一个像“BCGAYYY”这样的辅音词，如何在这种情况下只提供输出错误的字符A（因为A不是consonat），得到类似“BCGA”的输出+索引？

我有下面的方法wordValidation（）但是这提供了整个单词及其索引...

public abstract class Words {
    private String wordDetail;
    private String wordContent;

    public Words(String wordDetail, String wordContent) throws InvalidWordException{
    this.wordContent = wordContent;
    this.wordDetail = wordDetail;
    wordValidation();



    }

    public String getWordDetail() {
        return this.wordDetail;
    }

    public String getWordContent() {
        return this.wordContent;

    }

    public abstract String AcceptedCharacters();

    public void wordValidation() throws InvalidWordException{

          String content = getWordContent();
        String theseletters = this.AcceptedCharacters();


        for (int i = 0; i < content.length(); i++) {
            char c = content.charAt(i);
            if (theseletters.indexOf(c) == -1) {
                throw new InvalidWordException(content, i);

            }
        }

    }

    public String toString(){
    return  getWordDetail() + getWordContent();

    }

检查异常

public class InvalidWordException  extends Exception {

    public InvalidWordException (String wordContent, int theIndex) {
        super("Wrong Word" + wordContent + theIndex);
    }


}

具体等级1

public class Vowels extends Words {
    private String validVowels;

    public Vowels(String wordDetail, String wordContent) throws InvalidWordException {
        super(wordDetail, wordContent);
    }

    @Override
    public String AcceptedCharacters() {
        return validVowels = "AEIOU";
    }

    public static void main(String[] args) {
        try {
            Vowels vowel = new Vowels("First Vowel Check" ,"AEIOXAEI");
        } catch (InvalidWordException ex) {
            System.out.println(ex.getMessage());
        }
    }

}

具体课程2

public class Consonants extends Words {
  private String validConsonants;
    public Consonants(String wordDetail, String wordContent) throws InvalidWordException{
        super(wordDetail, wordContent);
    }

    @Override
    public String AcceptedCharacters() {
        return validConsonants ="BCDFGHJKLMNPQRSTVXZWY";
    }
    public static void main(String[] args) {
      try {
          Consonants consonants = new Consonants("First Consonant Check","BCGAYYY");
      } catch (InvalidWordException ex) {
          System.out.println(ex.getMessage());
      }

    }
}

测试程序

   public static void main(String[] args) {
      try {
          Consonants consonants = new Consonants("First Consonant Check","BCGAYYY");
      } catch (InvalidWordException ex) {
          System.out.println(ex.getMessage());
      }

    }

Answer 1

更改throw new InvalidWordException(content, i);

到

throw new InvalidWordException(content.substring(0,i), i);

在Java中，String对象是不可变的。所以你按原样传递原始内容字符串。这就是为什么它没有给你你想要的输出。

仅将字符串内容返回到特定索引

1 个答案: