Question

所以我的问题是，当我尝试并捕获输入错误时，我不知道如何继续我的程序。我尝试过使用“继续”;我的catch语句之后的代码，但这只是无法控制地循环我的程序。我需要程序在用户输入错误后从中断处开始。任何帮助，将不胜感激。请注意，这是一项任务，但我已经超越了处理代码中的垃圾。

      //Import library
      import java.io.*;
      import java.util.*;

      //File name
   public class GuessingGame
   {

//Main throws Input and output error
public static void main (String [] args) throws IOException
{
    //inputs for users
    Scanner in = new Scanner (System.in);
    Scanner i = new Scanner (System.in);
    //variables for the loop, random number, character and counter
    int guess = 0;
    int rnd;
    char decision;
    boolean loop = false;
    //random number generator
    Random random = new Random();
    rnd = random.nextInt(100) + 1;

    //loops the guess and input
    while (!loop){
        try{
            System.out.println(rnd);
            //prompt the user
            System.out.println(" Please guess a number between 1-100. Press 0 to exit.");
            int num = in.nextInt();
            //if statements

            if (num==0) 
            {
                //when user types '0' it ends the program
                System.exit(0);
                System.out.println("You gave up!.... Reseting program...");
            }
            else if (num>rnd) 
            {
                //prints too big, adds to counter 'guess'
                System.out.println("The number is too big!"); 
                guess++;
            }
            else if (num<rnd)
            {
                //prints too small, adds to counter 'guess'
                System.out.println("The number is too small!"); 
                guess++;
            }
            else 
            {
                //prints correct, adds to counter, dsiplays # of guesses and ends loop
                System.out.println("You guessed the number right!!"); 
                guess++; 
                System.out.print(" # of guesses: " + guess); 
                //Note**: i could've just put 'break;' but the compiler would'nt read the rest                       of the code below
                loop = true;
                //loops the case untill correct input is chosen either 'Y' or 'N'
                while(true){
                    //prompt the user if they want to play again
                    System.out.println(" Would you like to play again? Y/N?");
                    decision = i.nextLine().charAt(0);

                    switch (decision) {
                        case 'Y':
                        case 'y':    
                            //calls main, basically restarts the game
                            GuessingGame.main(args);     
                            break;

                        case 'N':
                        case 'n':
                            System.out.println("Bye!");
                            //exits the program completely
                            System.exit(0);
                            break;

                        default: 
                            //if incorrect input, this prints
                            System.out.println("Please enter a Yes or No <Y/N>");

                    }
                }
            }



        }
        //catches input errors
        catch (Exception e){ 
            System.out.println("Only numbers!!!");
            //GuessingGame.main(args);
            continue;

        } 
       }
       }

Answer 1

默认情况下，扫描程序将空格分割为标准输入，并保留已解析的子字符串数的索引。您调用的特定方法（.nextWhatever）将尝试将下一个字符串解析为其预期类型，并且只有在成功时才会增加索引;如果没有要解析的流，它将等待新的输入。

循环无限的原因是因为它无法将标记解析为整数并且没有增加索引。有两种方法可以跳过无效输入。 nextLine()将跳过剩余的流等待。例如，如果输入为“1 abc 2”

in.nextInt(); // equals 1
// in.nextInt() would fail
in.nextLine(); // equals "abc 2" and if put in your catch would clear the stream

但是，如果你想继续尝试后续令牌（在这种情况下跳过“abc”但尝试“2”，这是有效的），next()更合适，因为它只会跳过一个令牌。 / p>

try(){
  // validate input
  int num = in.nextInt();
}
catch(Exception e){
  System.out.println("Ignoring your faulty input");
  in.next();
}

Answer 2

尝试此举动，因为您只是测试输入。还要在你的catch中添加in.nextLine（）以消耗掉留下的角色。

while (!loop){
  int num;
     try{
                System.out.println(rnd);
                //prompt the user
                System.out.println(" Please guess a number between 1-100. Press 0 to exit.");
                num = in.nextInt();
            }
                catch (Exception e){ 
                    System.out.println("Only numbers!!!");
                    //GuessingGame.main(args);
                    in.nextLine();
                    continue;

                }

如何在Try Catch for Guessing游戏中继续循环

2 个答案: