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Java在“无效输入”上循环切换案例语句

作者:互联网

如果用户输入未定义选项的任何内容,我想更新代码以循环我的swtich语句.我在这里搜寻了许多网页,这些网页从各种搜索字词返回,并且相距很近,但到目前为止仍然不走运.
我的代码在这里应该可以使任何想要刺中它的人.

 java.util.Scanner;
 //import java.lang.Character.*; 
 //Thought this was needed to grab single char but its not
 public class caseloop {
 //main Method
 public static void main(String[] args)
 {
  Scanner input=new Scanner(System.in); //make so you can give input
  boolean go = true; // for starting main outer loop
  boolean run=true; // start inner loop
  while (go==true)
  {
    while (run==true)
    {
       //Output
       System.out.println("Enter option \n 1-Do this \n 2-Do this thing \n 3-Do this other thing");
       int option= input.nextInt(); //grab option number

       switch(option)
       {
         /*
          * This needs to loop and prompt user again if anything other than 1,2, or 3 is entered.
          */
         case 1:
           System.out.println("Option1");
           break;
         case 2:
           System.out.println("Option2");
           break;
         case 3:
           System.out.println("Option3");
           break;
         /*case 4:
           System.out.println("Option1");
           System.out.println("Option2");
           System.out.println("Option3");

           break;
         *
         * 
         * Case 4 was for debug
         * 
         */
         default:
           System.err.println("Invalid option selected");
           /*
            * On input that is not defined with in the switch-case it will revert to "default"
            * this fault staement needs to tell ther usere their option is not vaild and then
            * prompt them to try it again to enter an option. I can not get it to reprompt. 
            * I have tried a while and an if loop both sorta worked but did not actually loop
            * back to display again. I have been instucted that I am to not use a  try catch statment 
            * unless of course that is the only viable option in whichcase I will use it anyways.
            */

           //stupid default statement and its redundent built in "break;"

       }
      run=false;
      }


   /*
    * Outer Loop to prompt user if they want to run the entire program again with new entries.
    */
   if (run == false) 
   {
     System.out.println("Would you like to run again? Y/N");
     char again = input.next().charAt(0);
     again = Character.toUpperCase(again); //force all leters inputed to upper case, lower would work too if i change if conditions
     if (again == 'Y')
     {
       run = true;
     }
     else if (again == 'N')
     {
       System.out.println("Goodbye.");
       go=false;
     }
     else
     {
       System.err.println("Invalid entry. Try again.");
     }
   }
  }
 }
  //System.err.println("An error occured please try again");

}

在这方面的任何帮助将不胜感激.

解决方法:

您以一种非常奇怪的方式使用运行变量.由于在循环结束时将运行设置为false,因此循环将永远不会重复.如果更改它,以便仅将有效选项设置为run = false,则输入错误的选项将导致循环再运行一次.

在switch语句的末尾删除run = false,将其添加到System.out.println(“ OptionX”);之后;

标签:loops,switch-statement,java
来源: https://codeday.me/bug/20191101/1985256.html