Java 我需要跳过代码中我不需要的一部分'；我不知道怎么做

我想知道如果我在“输入”中的输入无效，如何跳过重试部分。例如，我输入“o”作为我的选择。它应该显示“无效输入”，并且应该显示菜单部分（跳过重试）。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"));
    }

}

}

---

正如@Kevin所说，你可以试试这个。

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);

通过此操作，您将获得以下结果：

• 输入

  I

  重试

  Y

  将再次运行循环
• 输入

  I

  重试

  N

  将终止循环
• 输入

  P

  重试

  Y

  将再次运行循环（显示无效输入，但让用户选择继续）
• 输入

  P

  重试

  N

  将终止循环（显示无效输入，用户决定不继续）

---

类似于

if（inputWasValid）{…}

？我想我想说的是，如果我输入了无效的输入，它不会给用户重试的选项。它将显示--家庭作业--部分，用户将输入另一部分choice@Ms.Smoak以下问题都不值得接受吗？

    ...
    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);