我有这个问题,我创建了两个验证用户输入的方法。然后我试图在程序的其余部分运行之前让他们的输入进行验证。它不起作用,我找不到任何有用的东西。
我在另一个程序中以完全相同的方式完成了它,但它不适用于此程序。任何帮助将不胜感激。
(主要部分已被注释掉,因为我试图让它运行)...
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;
}
}
答案 0 :(得分:4)
在方法getInt()
中,这将导致无限循环:
boolean valid = false;
while(valid == false);
{
由于尾部分号:删除它。尾部半冒号使while
等效于:
while (valid == false) {}
这意味着永远不会执行预期的循环体,并且永远不会更改valid
的值。
答案 1 :(得分:1)
虽然您的问题已经解决,但我想指出一些关于您的代码的事情......
首先,你的代码有很多重复..
您的方法: - getIntWithinRange
只是将您的请求委托给另一种方法 getInt
,我认为这种方法毫无用处。
您在 getInt
方法中所拥有的一切,只需将其移至 getIntWithinRange
方法即可。这样您就不会创建{{ 1}}变量,2 boolean
以及许多重复的代码..
此外,您不需要检查布尔值,如: -
2 while loops
此外,您可以将while (valid == false) // Not needed
while (!valid) // is enough
作为您的实例变量..您不需要在阅读用户输入的所有方法中定义它。实际上您不是..您只是复制它..
在评论的代码中: -
Scanner sc = new Scanner(System.in);
相当于: -
if (tries >= 7 && tries <=7)
在您的主要方法中: -
if (tries == 7)
在您的getInt方法中: -
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();