猜數字遊戲java

Question

我正在寫一個java代碼，要求用戶猜測計算機的隨機數，我很好，但是，我想要做的是有一個消息對應的猜測數拿走了用戶。

例如，如果用戶猜測1中的隨機數，嘗試輸出「優秀！」。如果用戶猜到2-4的答案嘗試「好工作」等。

我想我還沒有嘗試過任何東西，因爲我不知道在哪裏粘住代碼？
我知道它必須如果猜== = 1然後做這個如果代碼> 1 < = 3然後這樣做等等..但在我的代碼？它應該在嵌套if的while循環中嗎？

這是我更新的代碼，它運行並編譯我創建它所需的確切方式。感謝所有的幫助！

public static void main(String[] args) { 
     Random rand = new Random(); 
     int compNum = rand.nextInt(100); 
     int count = 0; 
     Scanner keyboard = new Scanner(System.in); 
     int userGuess; 
     boolean win = false;   
     while (win == false){ 
      System.out.print("Enter a guess between 1 and 100: "); 
      userGuess = keyboard.nextInt(); 
      count++; 
      if(userGuess < 1 || userGuess > 100){ 
       System.out.println("Your guess is out of range. Pick a number between 1 and 100."); 
       System.out.println(); 
      } 
      else if (userGuess == compNum){ 
       win = true; 
       System.out.println("Congratulations! Your answer was correct! "); 
      }  
      else if (userGuess < compNum){ 
       System.out.println("Your guess was too low. Try again. "); 
       System.out.println(); 
      } 
      else if (userGuess > compNum){ 
       System.out.println("Your guess was too high. Try again. "); 
       System.out.println(); 
      } 

     } 
     if(count == 1){ 
      System.out.println(); 
      System.out.println("I had chosen " + compNum + " as the target number."); 
      System.out.println("You guessed it in " + count + " tries."); 
      System.out.println("That was lucky!"); 
     } 
     else if (count > 1 && count <= 4){ 
      System.out.println(); 
      System.out.println("I had chosen " + compNum + " as the target number."); 
      System.out.println("You guessed it in " + count + " tries."); 
      System.out.println("That was amazing!"); 
     } 
     else if (count > 4 && count <= 6){ 
      System.out.println(); 
      System.out.println("I had chosen " + compNum + " as the target number."); 
      System.out.println("You guessed it in " + count + " tries."); 
      System.out.println("That was good."); 
     } 
     else if (count > 6 && count <= 7){ 
      System.out.println(); 
      System.out.println("I had chosen " + compNum + " as the target number."); 
      System.out.println("You guessed it in " + count + " tries."); 
      System.out.println("That was OK."); 
     }    
     else if (count > 7 && count <= 9){ 
      System.out.println(); 
      System.out.println("I had chosen " + compNum + " as the target number."); 
      System.out.println("You guessed it in " + count + " tries."); 
      System.out.println("That was not very good."); 
     } 
     else if (count > 10){ 
      System.out.println(); 
      System.out.println("I had chosen " + compNum + " as the target number."); 
      System.out.println("You guessed it in " + count + " tries."); 
      System.out.println("This just isn't your game."); 

     } 
    } 
     } 

Answer 1

保持一個計數的int i;，每個用戶猜測時間錯誤（如果沒有用戶得到正確的答案，做i++然後，有這樣的事，你迴應TI正確的答案如下：

if(i == 0) System.out.println("Perfect!"); //Got it the first time 
else if(i == 1) System.out.println("Nice!"); //Got it on the second try 
else if(i <= 4) System.out.println("Good Job!"); //Got it between the second and fifth time 
else if(i <= 8) System.out.println("Okay!"); //Got it between 6th and 9th time 
else System.out.println("Hmmm... Try to do better next time!"); //took more than 10 times to get it right 

Answer 2

你可以指望的迭代。喜歡的東西。

boolean notCorrect = true; 
int guesses = 0; 
while(notCorrect){ 
    //Code for checking user input. 
    //break out if true 
    guesses++; 
} 

對不起，這倒退了。

別處

if(guesses < 2) { 
    // display message 
} 
// include other if's to determine which message to display. 

你可以把if語句來決定在顯示哪一條消息如果檢查的猜測是否正確。但是把它拉出循環。這樣，如果用戶實際猜測正確，那麼您只能運行該代碼。

if (userGuess == compNum){ 
    win = true; 
    System.out.println("Congratulations! Your answer was correct! "); 
    // put message decision here... 
}