无法调用Java中的验证方法

Question

我有这个问题，我创建了两个验证用户输入的方法。然后我试图在程序运行的其余部分之前让他们的输入验证。这将无法正常工作，我无法在网上找到任何有助于解决问题的事情。

我在另一个程序中以完全相同的方式完成了它，但它在这个程序中不起作用。任何帮助将非常感激。

（主要部分被注释掉了，因为我的努力得到它运行）...

import java.util.*; 




public class GuessingGame { 

/** 
* @param args 
*/ 
public static void main(String[] args) { 
    // TODO Auto-generated method stub 



    int answer; 
    int tries = 0; 
    answer = (int) (Math.random() * 99 + 1); 



    System.out.println("Welcome to the Guess the Number Game "); 
    System.out.println("+++++++++++++++++++++++++++++++++++++ \n"); 
    System.out.println("I'm thinking of a number between 1-100 "); 
    System.out.println("Try to guess it! "); 

    Scanner sc = new Scanner(System.in); 
    String choice = "y"; 

while (choice.equalsIgnoreCase("y"))  
{ 



     int guess = getIntWithinRange(sc, "Enter number: ", 0, 100); 




     /** 

     if (guess == answer) 
     { 
      System.out.println("Your guess is correct! Congratulations!"); 
     } 
     else if (guess > answer + 10) 
      { System.out.println("Your guess was way too high"); 
      tries++; 
      } 

     else if (guess < answer) 
      { System.out.println("Your guess was too low. Try again. "); 
      tries++; 
      } 

     else if (guess > answer) 
      {System.out.println("Your guess was too high. Try again."); 
      tries++; 
      } 



       System.out.println("The number was " + answer + " !"); 
       System.out.println("You guessed it in " + tries + " tries"); 

        if (tries < 2) 
         {System.out.println("That was lucky!"); 
         } 

        if (tries >=2 && tries <=4) 
         {System.out.println("That was amazing!"); 
         } 

        if (tries > 4 && tries <= 6) 
         {System.out.println("That was good."); 
         } 

        if (tries >= 7 && tries <=7) 
         {System.out.println("That was OK. "); 
         } 

        if (tries > 7 && tries < 10) 
         { System.out.println("That was not very good. "); 
         } 

        if (tries >= 10) 
         {System.out.println("This just isn't your game. "); 
         } 
       **/ 

         //ask if they want to continue 
        System.out.println("\n Continue (y/n)? "); 
        choice = sc.next(); 
        sc.nextLine(); 
        System.out.println(); 

    } 

    //print out thank you message 
    System.out.println("Thanks for finding the common divisor "); 
    sc.close(); 
} 

public static int getInt(Scanner sc, String prompt) 
{ 
    int i = 0; 
    boolean valid = false; 

    while(valid == false); 
    { 
     System.out.println(prompt); 
     if (sc.hasNextInt()) 
     { 
      i = sc.nextInt(); 
      valid = true; 
     } 
     else 
     { 
      System.out.println("Please enter a number... "); 
     } 
     sc.nextLine(); 
    } 
    return i; 

} 








public static int getIntWithinRange(Scanner sc, String prompt, int min, int max) 
{ 
    int i = 0; 
    boolean valid = false; 

    while (valid == false) 
    { 
     i = getInt(sc, prompt); 
     if (i <= min || i >= max) 
      System.out.println("Number must be between 1-100 "); 
     else 
      valid = true; 
    } 
    return i; 
} 

}

Answer 1

在该方法中getInt()，这将导致一个无限循环：

boolean valid = false; 

while(valid == false); 
{ 

由于尾随分号：将其删除。拖尾半结肠使得while等效于：

while (valid == false) {} 

这意味着预期的循环体从不执行和的valid值没有改变。

Answer 2

虽然你的问题已经解决了，我想指出你的代码的一些东西..

首先，你的代码有那么多duplicacy的..

你的方法： - getIntWithinRange只是委托您的要求的另一种方法getInt，我认为这是没用..
你有getInt方法都什么，你可以简单地将它移动到getIntWithinRange方法里，你会不会创造2 boolean变量，2 while loops，并有许多重复的代码..

而且这样的话，就不需要检查布尔值，如： -

while (valid == false) // Not needed 
while (!valid)  // is enough 

此外，你可以让你的Scanner sc = new Scanner(System.in);作为你的实例变量..你不需要在你正在阅读用户输入的所有方法中定义它。而实际上你不是..你只是复制它..

在您所评论的代码： -

if (tries >= 7 && tries <=7) 

相当于： -

if (tries == 7) 

在main方法： -

System.out.println("\n Continue (y/n)? "); 
choice = sc.next(); 
sc.nextLine(); --> // You don't need this line at all.. 
        // It is just used to read user input.. 
        // That you are doing in your `getIntWithinRange` method.. 
System.out.println(); 

在你getInt方法： -

else { 
      System.out.println("Please enter a number... "); 
    } 
    sc.nextLine(); --> // This should be sc.next().. And should be inside else 
         // You are just getting a new user input.. not required here.. 
         // Just move the pointer to next input.. But don't read it..