java - 无法实现 java hangman 返回带有正确字母的空白(例如 "_ _ r _")

Question

我一直在为一个学校项目使用 Java BlueJ 开发一个基于文本的刽子手游戏，并且大部分都取得了成功，但是我在这部分遇到了一些困难。

我正在尝试做到这一点，以便每次玩家猜对一个字母时，它都会打印出空格(星号)并插入正确的字母。现在，我让它打印出你猜到的字母，并用正确的空格数打印出正确的猜测。 我现在唯一的问题是它在游戏开始时打印的空白数量不正确。

如何解决这个空格问题？

import java.util.*;

public class Hangman {
    Word w = new Word();
    private String word = w.chooseWord();
    private int count = 0;
    String guess;

public String getWord() {
    String w = word;
    return w;
}

public int countLetters() {
    for (int i = 0; i < word.length(); i++) {
        if (Character.isLetter(word.charAt(i))) {
            count++;
        }
    }
    return count;
}

public String Blanks() {
    countLetters();
    int num = 0;
    String spaces = "";
    String blankSpace = "*";
    while (num < count) {
        spaces = spaces + blankSpace;
        num++;
    }
    return spaces;
}

String wordSoFar = Blanks();

public int countOccurrences() {
    int count = 0;
    for (int i = 0; i < word.length(); i++) {
        if (word.substring(i, i + 1).equals(guess)) {
            count++;
        }
    }
    return count;
}

public String wordSoFar() {
    for (int i = 0; i < word.length(); i++) {
        if (word.substring(i, i + 1).equals(guess)) {
            wordSoFar = wordSoFar.substring(0, i) + guess + wordSoFar.substring(i + 1 , wordSoFar.length());
        }
    }
    return wordSoFar;
}  

public void Guess() {
    //Removed code that draws hangman due to it making this really long

    boolean correct = false;
    Scanner scan = new Scanner(System.in);
    int numIncorrectGuesses = 0;
    int numCorrectGuesses = 0;

    while (numCorrectGuesses != word.length() && numIncorrectGuesses < 6) {  
        guess = scan.next().substring(0,1);

        if (word.contains(guess)) {
            correct = true;
            numCorrectGuesses += countOccurrences();
            System.out.println(wordSoFar());
        } else {
            correct = false;
            numIncorrectGuesses++;
        }

    //Removed code that draws hangman due to it making this really long

    if (numCorrectGuesses == word.length()) {
        System.out.println("You win");
    } else {
        System.out.println("You lose");
    }
}

添加驱动类:

public class runGame {
    public static void main(String[]args) {
        Hangman game = new Hangman();
        System.out.println(game.getWord());
        System.out.println(game.Blanks());
        game.Guess();
    }
}

选择单词的类:

import java.util.*;

    public class Word {
        //String[] bank = {removing word bank due to length};

        public String chooseWord() {
            Random r = new Random();
            return new String(bank[r.nextInt(bank.length)]);
        }
    }

这是一个疯狂的例子(第一行只是为了测试；最终游戏不会显示这个词。单字符行是我的猜测):

object
************
    _______
   |/      |
   |
   |
   |
   |
   |
___|___
o
o*****
b
ob****
j
obj***
e
obje**
c
objec*
t
object
You win

Answer 1

你必须删除这条线

wordSoFar = Blanks();

从 wordSoFar() 函数开始。而是在 guess() 函数的开头执行。每次你猜一个新字符时，你都会用空白初始化它。每次调用 Blank() 函数时，它都会增加计数(因为它是一个类变量)，最终会增加“_”的数量。

现在针对您当前的双 * 问题，在 Hangman 类中为 wordSoFar 编写一个 getter。

public String getWordSoFar() {
    return wordSoFar
}

在主函数中

public class runGame {
    public static void main(String[]args) {
        Hangman game = new Hangman();
        System.out.println(game.getWord());
        System.out.println(game.getWordSoFar());
        game.Guess();
    }
}

还有一个想法，countLetter() 函数应该在计数之前将计数初始化为 0。

public int countLetters() {
    int count = 0;
    for (int i = 0; i < word.length(); i++) {
        if (Character.isLetter(word.charAt(i))) {
            count++;
        }
    }
    return count;
}