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