Java猜測遊戲調試問題我無法解決

Question

好吧，所以我的猜測遊戲幾乎完成我只是不知道爲什麼中間部分不工作。基本上，用戶輸入一個五位數的數字，它應該在他們的猜測中加入數字，然後增加他們的總數，並告訴他們有多少數字他們得到正確，所以他們可以再次猜測，但它甚至沒有通過我的循環任何幫助，將不勝感激。

//Jonathan 
import javax.swing.JOptionPane; 

import java.util.Scanner; 
public class Guessing 
{ 

public static void main(String[] args) 
{ 
    //JOptionPane.showMessageDialog(null, "Welcome to Jonathan's number guessing game."); 
    int correctPosition = 0, sum = 0, numOfGuesses = 5, i; 
    String usersGuess, secretCode; 
    double number = Math.random() * 100000; 
    number = Math.round(number); 
    secretCode = Double.toString(number); 
    do 
    { 
    numOfGuesses--; 
    System.out.println(number); 
    Scanner keyboard = new Scanner(System.in); 
    usersGuess = keyboard.nextLine(); 
    if (usersGuess.equalsIgnoreCase(secretCode)) 
    { 
     JOptionPane.showMessageDialog(null, "You got the number correct!"); 
     System.exit(0); 
    } 
    else if (!(usersGuess.equalsIgnoreCase(secretCode))) 
    { 
    for (i=0; i > 5; i++) 
     { 
      int digit = Character.getNumericValue(usersGuess.charAt(i)); 
      if (digit == secretCode.charAt(i)) 
      System.out.println(digit); 
      { 
       correctPosition++; 
      } 

     } 
     JOptionPane.showMessageDialog(null, "You have " + correctPosition + " numbers in the correct position." + " The sum of which are\n" + sum + " keep guessing!"); 
    } 
    } while (numOfGuesses > 0); 

    } 
} 

Answer 1

幾件事情中脫穎而出......

首先...

for (i = 0; i > 5; i++) { 

當i爲0，i將小於5，這意味着i > 5將false和環將終止...

我想你的意思是

for (i = 0; i < secretCode.length(); i++) { 

如果事實，因爲你需要在secretCode和userGuess每個字符比較，你真的需要一個複合環

for (i = 0; i < secretCode.length(); i++) { 
    for (int x = 0; x < usersGuess.length(); x++) { 
     //... 
    } 
} 

接下來，你的usersGuess值轉換爲int，但是比較它的secretCode的char值...

int digit = Character.getNumericValue(usersGuess.charAt(i)); 
if (digit == secretCode.charAt(i)) 

這將永遠是平等的，相反，簡單地比較userGuess的char值對的secretCode

for (i = 0; i < secretCode.length(); i++) { 
    for (int x = 0; x < usersGuess.length(); x++) { 
     char digit = usersGuess.charAt(x); 
     if (digit == secretCode.charAt(i)) { 
      System.out.println(digit); 
      correctPosition++; 
     } 
    } 
} 

現在char值，這可能會產生假postives，更多的東西一樣......

for (i = 0; i < secretCode.length(); i++) { 
    if (i < usersGuess.length()) { 
     char digit = usersGuess.charAt(i); 
     if (digit == secretCode.charAt(i)) { 
      System.out.println(digit); 
      correctPosition++; 
     } 
    } 
} 

威力比較合適，但我也會添加一個if聲明來檢查usersGuess的length，以確保它等於的length並顯示錯誤如果用戶輸入無效值......這是我一直在做... 和...

if (digit == secretCode.charAt(i)) 
System.out.println(digit); 
{ 
    correctPosition++; 
} 

這是因爲實際執行[R消息...

if (digit == secretCode.charAt(i)) { 
    System.out.println(digit); 
} 
correctPosition++; 

哪個我不牛逼認爲這是你想要什麼，我想你想要的東西更像

if (digit == secretCode.charAt(i)) { 
    System.out.println(digit); 
    correctPosition++; 
}