带有switch语句的Java嵌套循环问题

Question

对于案例“1”，它似乎只是循环而不是跳转到drunkSong();。如果用户选择案例“2”，它将正常工作并要求用户输入数字以返回 int“BeerNum”。

案例“1”需要接受默认值 99，而案例“2”需要接受用户输入。

我附上了代码，如果你能指出我的错误或我哪里出错了。

package PartOne;
import java.util.Scanner;

public class View {

    public void begin() {
        BeerSong.drunkSong();
    }

    private int setBeerNum() {
        return beerNum;
    }

    private int beerNum;

    public Integer Menu() {
        Scanner in = new Scanner(System.in);
        // print menu
        for (int i = 1; i <= 3; i++)
            System.out.println(i + ". Menu item #" + i);

        System.out.println("0. Quit");
        // handle user commands
        boolean quit = false;
        int menuItem;
        do {
            System.out.print("Choose menu item: ");
            menuItem = in.nextInt();
            switch (menuItem) {
                case 1:
                    System.out.println("Default ");
                    begin();
                    while (true)
                        try {
                            //where I have made a mistake
                            beerNum = 99;
                            this.beerNum = setBeerNum();
                            break;

                        } catch (NumberFormatException BecauseIsaidSo) {
                            System.out.print("Try again: ");
                        }

                    //working
                case 2:
                    System.out.println("Enter your number to play the song: ");
                    Scanner scan = new Scanner(System.in);
                    while (true)
                        try {
                            beerNum = Integer.parseInt(scan.nextLine());
                            this.beerNum =  setBeerNum();
                            return beerNum;

                        } catch (NumberFormatException BecauseIsaidSo) {
                            System.out.print("Try again: ");
                        }
                case 0:
                    quit = true;
                    break;

                default:
                    System.out.println("Invalid choice.");
            }
        }
        while (!quit) ;
        System.out.println("Bye-bye!");
        return null;
    }
}

Answer 1

我认为将菜单逻辑封装在单独的类中并将所有逻辑封装在其中会更好。此外，每个菜单项的逻辑也应该封装在单独的方法中。

public final class Menu {

    private Integer res;
    private boolean quit;

    public static Integer showMenuAndGetResult() {
        Menu menu = new Menu();
        menu.activate();
        return menu.res;
    }

    private Menu() {
    }

    private void activate() {
        try (Scanner in = new Scanner(System.in)) {
            show();

            while (!quit) {
                System.out.print("Choose menu item: ");
                int menuItem = in.nextInt();

                if (menuItem == 1)
                    onMenuItem1(in);
                else if (menuItem == 2)
                    onMenuItem2(in);
                else if (menuItem == 0)
                    onMenuQuit();
                else
                    System.out.println("Invalid choice.");
            }

            System.out.println("Bye-bye!");
        }
    }

    private void show() {
        for (int i = 1; i <= 3; i++)
            System.out.println(i + ". Menu item #" + i);

        System.out.println("0. Quit");
        System.out.println();
    }

    private void onMenuItem1(Scanner scan) {
        System.out.println("Default ");
        // TODO menu 1 logic incapsulation
    }

    private void onMenuItem2(Scanner scan) {
        System.out.println("Enter your number to play the song: ");
        // TODO menu 2 logic incapsulation
    }

    private void onMenuQuit() {
        res = null;
        quit = true;
    }

}

P.S.我个人尽量避免使用switch。这不是我的建议，请注意。我认为 if...else if 更具可读性和无问题。