2014-12-10 31 views
0

我試圖在Visual C#中創建一個Hang子手遊戲,但我遇到了按鈕單擊事件的信函問題。Visual C#Hang子手按鈕單擊事件

這是它應該如何工作的:玩家點擊一個字母按鈕(A-Z)。如果該按鈕的文本位於他應該猜測的單詞(guessWord)中,則該字母被顯示並且正確的猜測計數器遞增。如果這封信不在猜測中,他留下的機會就會減少,而一名hang子手也會被吸引。當guessGuess> = guessWord的長度時,玩家贏得比賽,當他沒有機會時(他以8開始),他將失去比賽。

我不知道我現在的代碼有什麼問題。有時當我運行它時,在點擊一個字母按鈕後,我會看到「你丟失了」框。有時,當一個正確的字母被猜出時,程序會顯示該字母,但會顯示「You lost」框。

void LetterBtn_Click(object sender, EventArgs e) 
{ 

    //Event hook-up 
    Button b = (Button)sender; 
    char letterClicked = b.Text.ToCharArray()[0]; 

    //Disable button after it's been clicked 
    b.Enabled = false; 

    string gameOverTitle1 = "Congrats!"; 
    string gameOverTitle2 = "Sorry!"; 
    string gameOverMsg1 = "You won!"; 
    string gameOverMsg2 = "You lost!"; 

    char[] guessWordChars = guessWord.ToCharArray(); 

    //If the character is in the word 
    if ((guessWord = guessWord.ToUpper()).Contains(letterClicked)) 
    { 

     for (int i = 0; i < guessWord.Length; i++) 
     { 

      if (guessWordChars[i] == letterClicked) 
      { 

       //Reveal the character 
       lblguessWord[i].Font = new Font("Bauhaus 93", 28, FontStyle.Regular); 
       lblguessWord[i].Text = letterClicked.ToString(); 
       lblguessWord[i].ForeColor = Color.Fuchsia; 
       //Increment counter 
       correctGuess += 1; 

      } 

     } 

     //Winning condition 
     if (correctGuess >= guessWordChars.Length) 
     { 
      return; 

     } 

     //Tell the player he won the game 
     frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle2, gameOverMsg2); 
     frmGameOverBoxInstance.ShowDialog(); 
    } 

    else 
    { 

     //Incorrect guess 
     if (chances > 0) 
     { 

      chances -= 1; 
     } 

     //Player lost the game 
     else 
     { 

      //Reveal the word 
      for (int k = 0; k < guessWordChars.Length; k++) 
      { 

       lblguessWord[k].Text = guessWordChars[k].ToString(); 

      } 

      //Tell the player he lost the game 
      frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle1, gameOverMsg1); 
      frmGameOverBoxInstance.ShowDialog(); 
     } 
    } 
} 

回答

2

你的問題是在這裏:

//Tell the player he won the game 
frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle2, gameOverMsg2); 
frmGameOverBoxInstance.ShowDialog(); 

你是混合了自己的標籤。 gameOverMsg2包含您的You Lost!字符串。將其更改爲這個:

//Tell the player he won the game 
frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle1, gameOverMsg1); 
frmGameOverBoxInstance.ShowDialog(); 

這裏,您有同樣的問題,在您的else條款:

//Tell the player he lost the game 
frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle1, gameOverMsg1); 
frmGameOverBoxInstance.ShowDialog(); 

更改它以這樣的:

//Tell the player he lost the game 
frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle2, gameOverMsg2); 
frmGameOverBoxInstance.ShowDialog(); 

你應該給你的變量名更多的描述性標題來避免這種情況,例如gameOverLostMsg而不是gameOverMsg2gameOverWinMsg而不是gameOverMsg1

1

您在這裏有一個問題:

//Winning condition 
    if (correctGuess >= guessWordChars.Length) 
    { 
     return; 

    } 

    //Tell the player he won the game 
    frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle2, gameOverMsg2); 
    frmGameOverBoxInstance.ShowDialog(); 

此代碼返回如果用戶猜中所有的字母,併爲用戶提供獲勝的消息,如果他/她猜,但還沒有贏了。你可能想做到以下幾點來代替:

 //Winning condition 
     if (correctGuess >= guessWordChars.Length) 
     { 
      //Tell the player he won the game 
      frmGameOverBox frmGameOverBoxInstance = new frmGameOverBox(gameOverTitle2, gameOverMsg2); 
      frmGameOverBoxInstance.ShowDialog(); 
     } 

除此之外,在開始添加一些調試代碼,以確保您的變量是正確的初始化,像這樣:

Debug.WriteLine("correctGuess: " + correctGuess.ToString()); 
    Debug.WriteLine("chances: " + chances.ToString()); 
+1

除了這一切,如果有人猜測已經被猜出並已被揭示的信件,沒有什麼可以阻止他們再次猜測同一封信。更糟糕的是,那個人可以不斷猜測同一封信,每次都會被視爲一個正確的猜測,從而最終滿足'如果'條件並最終贏得每場比賽。但不是真的。 – Icemanind 2014-12-10 23:19:01

+0

@icemanind - 是的,@ mdegges還應該保留一個布爾型'猜測'標誌數組,以防止這種情況。 – dbc 2014-12-10 23:23:44

+0

b.Enabled = false;阻止用戶多次點擊一個字母按鈕。 – mdegges 2014-12-10 23:27:14