java - 如何退出java方法

Question

抱歉，标题不好，我不知道更好的解释。

我目前正在编写一个杂货 list 程序，但我陷入了我的 addItemsToList 方法。基本上我希望它必须有“退出”点。其中之一将带您返回应用程序的主菜单，其中之一将带您返回该 addItemsToList 方法的上下文菜单。不幸的是我完全陷入困境。

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，我可以完美退出addItemsToList，但如果i等于“exit”，我似乎找不到退出的逻辑解决方案。我让它工作得又快又脏，应用程序首先询问数量，然后退出，但这并不是我想要的。

由于我(显然)是初学者，因此对我的风格或其他方面的任何其他评论也将不胜感激!!

Answer 1

您可以使用 switch case 来实现此目的，使用一个标志来了解何时执行优雅的退出，并使用 while 循环来帮助您留在菜单中，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，您可以拥有多个相互调用的菜单，最后，程序自行完成，无需疯狂返回；或 System.exit(0);

编辑:添加了一个类以获得更好的洞察力。