如何退出java方法

Question

抱歉标题不好，我不知道更好的解释..

所以我目前正在编写一个杂货清单程序，但我被我的 addItemsToList 方法困住了。基本上我希望它必须“退出”点。其中之一将带您回到应用程序的主菜单，其中之一将带您回到 addItemsToList 方法的上下文菜单。 不幸的是，我完全被卡住了。

这是我的方法代码：

 private static void addItemsToList(){
        System.out.println("To which category do you want to add? ");
        categoryInstructions();
        int choice2 = scan.nextInt();

        if(choice2 == 5){ // i am "getting out" here just fine
            interact(); // this is the method that instructs everything to happen in the main method
        }else {
            System.out.println("What do you want to add? Type 'exit' for menu");
                i = scan.next(); // this is where the second "exit" point is supposed to be if someone types "exit"

                if (i.compareTo(exit) != 0) {                                         
                        System.out.println("Please enter the quantity ");
                        int quant = scan.nextInt();

                        if (choice2 == 1) {
                            GrocerieList.foodstuffsList.add(i);
                            GrocerieList.foodstuffsAmount.add(quant);
                        } else if (choice2 == 2) {
                            GrocerieList.hygeneList.add(i);
                            GrocerieList.hygeneAmount.add(quant);
                        } else if (choice2 == 3) {
                            GrocerieList.drinkList.add(i);
                            GrocerieList.drinkAmount.add(quant);
                        } else if (choice2 == 4) {
                            GrocerieList.otherList.add(i);
                            GrocerieList.otherAmount.add(quant);
                        }
                    } else {
                        addItemsToList();
                }
        }
    }

为了更好地理解这里是 categoryInstructions()：

public static void categoryInstructions(){
        System.out.println("Press");
        System.out.println(" ");
        System.out.println("\t\t" + "(1)" + "\t--\t" + "for foodstuffs");
        System.out.println("\t\t" + "(2)" + "\t--\t" + "for hygene");
        System.out.println("\t\t" + "(3)" + "\t--\t" + "for drinks");
        System.out.println("\t\t" + "(4)" + "\t--\t" + "for others");
        System.out.println(" ");
        System.out.println("\t\t" + "(5)" + "\t--\t" + "to get back main menu");
    }

现在发生的事情是，如果choice2 == 5，我可以完全退出 addItemsToList，但如果我等于“退出”，我似乎无法找到退出的逻辑解决方案。我让它在应用程序中快速而肮脏地工作，首先询问数量然后退出，但这并不是我想要的。

由于我（显然）是初学者，因此我的风格中的任何其他 cmet 也将不胜感激！

Answer 1

为此，您可以使用 switch case、知道何时执行优雅退出的标志和帮助您留在菜单中的 while 循环，直到用户提供正确的输入。

例如：

public class Controller {

    private GroceriesController groceriesController;
    private boolean isRunning; // our flag that will help us to quit the menu

    public Controller(GroceriesController groceriesController) {
        this.groceriesController = groceriesController;
    }

    public void run() {
        isRunning = true;
        while (isRunning) {
            clearScreen();
            displayLogo();
            displayMainMenu();
            handleMainMenu(); // method that will do all the job;
        }
    }

    private void handleMainMenu() {
        int userChoice = scanner.nextInt();
        switch (userChoice) {
            case 1:
                groceriesController.handleGroceriesMenu();
                break;
            case 2:
                returnItems();
                break;
            case 0:
                isRunning = false; // exit from the menu
                break;
            default:
                System.out.println("Invalid input");
                break;
        }

    }

    private void displayMainMenu() {
        System.out.println("\n Grocery store:\n"
                + "[1] Buy stuff\n"
                + "[2] Return stuff\n"
                + "[0] Leave store");
    }
}

public class GroceriesController {

    private boolean isRunning; // same strategy as before

    public void handleGroceriesMenu() {
        isRunning = true;
        while (isRunning) {
            clearScreen();
            displayLogo();
            displayMenu();
            handleMenu();
        }
    }

    private void handleMenu() {
        int userChoice = scanner.nextInt();
        switch (userChoice) {
            case 1:
                addItemToCart(item);
                System.out.println("The item: " + item + "has been added.");
                break;
            case 2:
                removeItemFromCart(item);
                System.out.println("The item: " + item + "has been removed.");
                break;
            case 0:
                isRunning = false; // exit from the menu
                break;
            default:
                System.out.println("Invalid input");
                break;
        }

    }

    private void displayMenu() {
        System.out.println("\n Select:\n"
                + "[1] Buy item\n"
                + "[2] Put back\n"
                + "[0] Leave");
    }

}

通过这种方式，你可以有多个菜单相互调用，最后程序自行完成，没有狂野的返回；或 System.exit(0);

编辑：增加了一个类以获得更好的洞察力。