隐藏子菜单答案

【问题标题】：Hiding a submenu隐藏子菜单
【发布时间】：2016-02-23 07:38:39
【问题描述】：

当我尝试在 while 循环之外使用“exit”(8) if 语句时，如何防止显示子菜单？目前，当我输入 8 时，它会显示子菜单，然后说退出，并执行它所在的测试器类的其余功能。

    mainMenu.addOption("Add Item");  // creating menu
    mainMenu.addOption("Remove Item");
    mainMenu.addOption("");
    mainMenu.addOption("");
    mainMenu.addOption("");
    mainMenu.addOption("");
    mainMenu.addOption("");
    mainMenu.addOption("Exit");

    subMenu.addOption("Burger");
    subMenu.addOption("Drink");
    subMenu.addOption("Fries");

    int input;
    int choice;
    do{
        input = mainMenu.getInput();
        choice = subMenu.getInput();
        if(input == 1){
            if(choice == 1){
                System.out.println("Burger Added");
            }
            if(choice == 2){
                System.out.println("Drink Added");
            }
            if(choice == 3){
                System.out.println("Fries Added");
            }
        }
        if(input == 2){
            if(choice == 1){
                System.out.println("Burger Removed");
            }
            if(choice == 2){
                System.out.println("Drink Removed");
            }
            if(choice == 3){
                System.out.println("Fries Removed");
            }
        }
    } while(input != 8);
    if(input == 8){
        //choice = 8;
        System.out.println("exiting");
    }

这是菜单类

import java.util.ArrayList;
import java.util.Scanner;

/**
    A menu that is displayed on a console.
 */
public class Menu
{
    // Declare instance variables ****
    // Global
    private ArrayList<String> options;


private Scanner in;

   /**
      Constructs a menu with no options.
   */
   public Menu()
   {
       /* Constructor - a special method whose name matches the class name exactly. Takes no or many arguments
       *  no return type for constructors
       *  purpose is to initialize data variables
       */
      options = new ArrayList<String>();
      in = new Scanner(System.in);
   }


   /**
      Adds an option to the end of this menu.
      @param option the option to add
   */
   public void addOption(String option)
   {
      options.add(option);
   }

   /**
      Displays the menu, with options numbered starting with 1,
      and prompts the user for input. Repeats until a valid input
      is supplied.
      @return the number that the user supplied
   */
   public int getInput()
   {
      int input;
      do
      {
         for (int i = 0; i < options.size(); i++)
         {
            int choice = i + 1;
            System.out.println(choice + ") " + options.get(i));
         }
         input = in.nextInt();
      }
      while (input < 1 || input > options.size());
      return input;
   }
}

【问题讨论】：

您应该研究if-then-else 语句及其相对于使用多个if-then 语句的好处（在某些情况下）。 The if-then and if-then-else Statements
与问题无关
如果您仔细研究过它，您应该会看到另一种编写程序的方式。您的问题的答案：使用while 循环而不是do-while 并让它检查input != 8。无论条件如何，do-while 循环至少迭代一次。
不会导致无限循环
我会发布一个答案以更清楚地显示它。

标签： java menu submenu display

【解决方案1】：

以下代码将首先在while 循环之外收集输入，如果input == 8 则退出，否则将通过子菜单。最后再次收集输入。

input = mainMenu.getInput();

while (input != 8) {

    choice = subMenu.getInput();

    if (input == 1) {
        if (choice == 1) {
            System.out.println("Burger Added");
        } else if (choice == 2) {
            System.out.println("Drink Added");
        } else if (choice == 3) {
            System.out.println("Fries Added");
        }
    } else if (input == 2) {
        if (choice == 1) {
            System.out.println("Burger Removed");
        } else if (choice == 2) {
            System.out.println("Drink Removed");
        } else if (choice == 3) {
            System.out.println("Fries Removed");
        }
    }

    input = mainMenu.getInput();
}

if (input == 8) {
    System.out.println("exiting");
}

下一个do-while 循环将在循环内收集输入，如果input == 8 则退出。

do {
    input = mainMenu.getInput();

    if (input != 8) {
        choice = subMenu.getInput();

        if (input == 1) {
            if (choice == 1) {
                System.out.println("Burger Added");
            } else if (choice == 2) {
                System.out.println("Drink Added");
            } else if (choice == 3) {
                System.out.println("Fries Added");
            }
        } else if (input == 2) {
            if (choice == 1) {
                System.out.println("Burger Removed");
            } else if (choice == 2) {
                System.out.println("Drink Removed");
            } else if (choice == 3) {
                System.out.println("Fries Removed");
            }
        }
    }

} while (input != 8);

if (input == 8) {
    System.out.println("exiting");
}

【讨论】：