Question

我按z时的错误是：

Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:864)
at java.util.Scanner.next(Scanner.java:1485)
at java.util.Scanner.nextInt(Scanner.java:2117)
at java.util.Scanner.nextInt(Scanner.java:2076)
at InputSum.main(InputSum.java:60)

这是我的计划：

import java.util.Scanner;

public class InputSum
{
   public static void main(String[] args)
   {
      Scanner input = new Scanner(System.in);

   //declarations
      int sum;
      int number;
      int letter;
      boolean cont = true;
      boolean mop = true;

   //outputs

      System.out.println("Input ONLY Positive Integers including zero to Add (end with -1): ");


   //variables
      sum = 0;
      String numbers = "";

   //logic

      while(cont)  // loop
      {    
         number = input.nextInt();
         if(number == -1)
            break;
         // create a break for -1

         else if(number > -1)
         {sum = sum + number; numbers = numbers + number + ", "; } // summation of the numbers and string 

         else 
            System.out.println("Invalid input. Enter a valid input");
            // for non positive integers the output is invalid
      }



    //outputs and loops

      System.out.println("Entered numbers: " + numbers);
      System.out.println("The sum: " + sum);
      System.out.println("Would you like to continue with new inputs?(enter z for yes and any number other than 50 for no)");
      letter = input.nextInt();
      while(cont)
      {
         if(letter == 50){
            System.out.println("Input ONLY Positive Integers including zero to Add (end with -1): ");

            while(cont)  // loop
            {    
               number = input.nextInt();
               if(number == -1)
                  break;
               // create a break for -1

               else if(number > -1)
               {sum = sum + number; numbers = numbers + number + ", "; } // summation of the numbers and string 

               else 
                  System.out.println("Invalid input. Enter a valid input");
            // for non positive integers the output is invalid
            }

         }  
         else 

            System.exit(0); 
      }   // output and print statement for the sum

   }     
}

我不知道如何修复它，所以当我点击z时，它表示是，并且不会给我这些错误。我该如何解决？

以下是该计划的说明：

练习＃1

编写一个名为InputSum的程序，提示用户输入   正整数。程序应该接受整数直到   用户输入值-1（负一）。用户输入-1后，   程序应显示输入的数字，后跟其总和   如下所示。

您必须设计程序，以便询问用户是否需要   在每组条目之后继续新输入，结束程序   只有当他们没有用＆＃34;是＆＃34;。
回答问题时
请注意，在以下示例运行中，-1不是输出的一部分   显示＆＃34;输入的数字＆＃34;：
Input numbers to add (end with -1): 10  2  13  50  100  -1
Entered numbers: 10, 2, 13, 50, 100
The Sum: 175

Would you like to continue with new inputs? yes

Input numbers to add (end with -1): 1  2  3  4  5  6  7  8  9  10  -1
Entered numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
The Sum: 55

Would you like to continue with new inputs? yes

Input numbers to add (end with -1): 1  1  1  1  100  -1
Entered numbers: 1, 1, 1, 1, 100
The Sum: 104

Would you like to continue with new inputs? yes

Input numbers to add (end with -1): 0  0  0  0  0  100  -1
Entered numbers: 0, 0, 0, 0, 0, 100
The Sum: 100 

Would you like to continue with new inputs? no    Make sure the program validates each entered number before processing it as      
用户可以输入句子值以外的负数 -1。如果是负数不是-1，则打印输出＆＃34;无效输入＆＃34;并继续接受数字。腾出你的输出如上所示。＆＃34;＆＃34;

Answer 1

当您要求用户输入＆＃34; z＆＃34;时，您使用letter = input.nextInt();，而您想要使用letter = input.next()或letter = input.nextLine() ＆＃34; Z＆＃34;是一个字符串而不是整数。当然letter是一个整数，所以你不能在其中存储一个字符串，因此需要一个新的变量

简单的Java错误，初学者CS学生

练习＃1

1 个答案: