Question

我想知道如果我在'输入'中的输入无效，我怎么能跳过我的尝试再次尝试。例如，我输入'o'作为我的选择。它应显示“无效输入”，它应显示菜单部分（再次跳过尝试）。请帮帮我。

public class Menu {

    public static void MainMenu() {
        Part1 call1 = new Part1();
        Part2 call2 = new Part2();
        Part3 call3 = new Part3();
        Part4 call4 = new Part4();

        Scanner in = new Scanner(System.in);
        String yn = null;


        do {
            System.out.println("\t\t---HOMEWORK---");
            System.out.println("\tI  for PART 1");
            System.out.println("\tII  for PART 2");
            System.out.println("\tIII  for PART 3");
            System.out.println("\tIV  for PART 4");
            System.out.print("\tEnter input:     ");
            String input = in.next();

            do {
                switch (input) {

                    case "I":
                        call1.one();
                        break;

                    case "II":
                        call2.two();
                        break;

                    case "III":
                        call3.three();
                        break;

                    case "IV":
                        call4.four();
                        break;

                    case "V":
                        System.exit(0);
                        break;

                    default:
                        System.out.println("invalid input");
                        break;

                }

                System.out.print("try again? -Y- || -N-     :  ");
                yn = in.next();

            } while (yn.equalsIgnoreCase("y"));

        } while (yn.equalsIgnoreCase("n"));
    }

}

Answer 1

public class Menu {

public static void MainMenu() {
    Part1 call1 = new Part1();
    Part2 call2 = new Part2();
    Part3 call3 = new Part3();
    Part4 call4 = new Part4();

    boolean inputWasValid = false; 

    Scanner in = new Scanner(System.in);
    String yn = null;


    do {
        System.out.println("\t\t---HOMEWORK---");
        System.out.println("\tI  for PART 1");
        System.out.println("\tII  for PART 2");
        System.out.println("\tIII  for PART 3");
        System.out.println("\tIV  for PART 4");
        System.out.print("\tEnter input:     ");
        String input = in.next();

        do {
            switch (input) {

                case "I":
                    call1.one();
                    break;

                case "II":
                    call2.two();
                    break;

                case "III":
                    call3.three();
                    break;

                case "IV":
                    call4.four();
                    break;

                case "V":
                    System.exit(0);
                    break;

                default:
                    inputWasValid = true;
                    System.out.println("invalid input");
                    break;

            }
            if (inputWasValid) {
               break;
            }
            System.out.print("try again? -Y- || -N-     :  ");
            yn = in.next();

        } while (yn.equalsIgnoreCase("y"));

    } while (yn.equalsIgnoreCase("n"));
}

}

@Kevin说，你可以试试这个。

Answer 2

我的理解是你想要这样的东西：

    ...
    Scanner in = new Scanner(System.in);
    String yn = null;
    boolean retry;


    do {
        System.out.println("\t\t---HOMEWORK---");
        System.out.println("\tI  for PART 1");
        System.out.println("\tII  for PART 2");
        System.out.println("\tIII  for PART 3");
        System.out.println("\tIV  for PART 4");
        System.out.print("\tEnter input:     ");

        String input = in.next();
        retry = true;

        switch (input) {

            ...

            default:
                System.out.println("invalid input");
                break;
        }

        System.out.print("try again? -Y- || -N-     :  ");
        yn = in.next();

        // might want to do check & loop here to see if user enters just Y or N 

        if(retry && yn.equalsIgnoreCase("N")) retry = false;   

    } while (retry);

有了这个，你会得到以下结果：

输入I再试一次Y将再次循环播放
输入I再试一次N将终止循环
输入P再次尝试Y将再次循环播放（显示无效输入，但允许用户选择继续）
输入P再次尝试N将终止循环（显示无效输入，用户决定不继续）

我需要跳过我的代码中的一部分，我不知道如何

2 个答案: