java - 为什么这段代码不起作用？刽子手

Question

我正在制作刽子手游戏。一切正常，我已经准备好用于使游戏失败并为猜测提供 -1 的代码。虽然将它添加到 else 语句时它会重复等于单词的长度并且它也会给出一个猜测——即使它是正确的？我看不出代码有什么问题？我相信这是我的代码，但我猜错了，但没有放对，尽管我没有其他办法？

这是代码:

private class check implements ActionListener {
   public void actionPerformed(ActionEvent ae) {
      try {
         // Grabs the letter from the guessField and converts it into a char
         // which can be used to compare against the word.
         guess = guessField.getText();
         guessField.setText("");
         char guess2 = guess.charAt(0);

         // --------------------
         // Here is the guessing logic but it's currently
         // not working and you can not win since i haven't written code for
         // it yet. it's not selecting all the letters. for Example if
         // choosing A in a word such as Banana it only selects the first
         // a--------------------------- //
         String displaySecret = wordField.getText();
         if (displaySecret.equals("")) {/* case for fist execution */
            displaySecret = "";
            for (int i = 0; i < random.length(); i++)
               displaySecret += "_ ";
         }
         String newDisplaySecret = "";
         for (int v = 0; v < random.length(); v++) {
            if (guess2 == random.charAt(v)) {
               newDisplaySecret += random.charAt(v); // newly guessed
                                                     // character
            } else {
               newDisplaySecret += displaySecret.charAt(v); // old state
               guesses--;
               statusLabel.setText("Guesses left: " + guesses);
               missField.setText(missField.getText() + guess);
               if (guesses <= 0) {
                  JOptionPane.showMessageDialog(null,
                        "Game over! The word was: " + random);
                  guessField.setEditable(false);
                  wordField.setText("");
                  missField.setText("");
                  guesses = 7;
                  statusLabel.setText("Guesses left: " + guesses);
               }
            }
         }
         displaySecret = new String(newDisplaySecret);
         wordField.setText(displaySecret);
         if (displaySecret.equals(random)) {
            JOptionPane.showMessageDialog(null, "You Won! The Word was: "
                 + random);
            guesses = 7;
            statusLabel.setText("Guesses left: " + guesses);
            wordField.setText("");
            missField.setText("");
            guessField.setEditable(false);
         }
      } catch (Exception e) {
         System.out.println(e);
      }
   }
}

Answer 1

如果 random 是您的单词，您将遍历它的每个字符，然后检查每个字符是否与您对每个与猜测 a -1 不匹配的字符的猜测相匹配。

例如:单词是 Bananarama，您猜 n 您的第一个和第二个匹配项将转到 else 子句。然后 if 子句再次出现一次，然后是 else 等等。

你必须

1. 遍历所有字符，检查它们是否匹配
2. 如果发生匹配，替换字符并增加计数器
3. 检查正确字符的计数器是否等于0
4. 如果是，减少猜测

一些其他提示:在比较之前，您应该在输入和单词字符串上使用 .toLower() 以允许不区分大小写

一些示例代码:

int charsGuessedCorrectly;
for ( int i = 0; i < random.length( ); i++ )
{
    if ( random.charAt( i ) == guess )
    {
        charsGuessedCorrectly++;
        newDisplaySecret += random.charAt(v); // newly guessed
                                              // character
    }
}

if ( charsGuessedCorrectly == 0 )
{
    newDisplaySecret += displaySecret.charAt(v); // old state
    guesses--;
    statusLabel.setText("Guesses left: " + guesses);
    missField.setText(missField.getText() + guess);
    if (guesses <= 0) {
       JOptionPane.showMessageDialog(null,
           "Game over! The word was: " + random);
       guessField.setEditable(false);
       wordField.setText("");
       missField.setText("");
       guesses = 7;
       statusLabel.setText("Guesses left: " + guesses);
}