电影项目和对象阵列

Question

我一直在尝试编码电影院项目，但我无法弄清楚。当我运行程序时，它只会输出最后一个值进入固定套件，因此即使他们的年龄不正确，也可以让他面前的每个人进入电影院。我相信这与不是数组的人对象有关，但是我不确定该如何解决这个想法？

import java.util.Scanner;

public class Cinema 
{
    public static void main(String[] args)
    {
        QueueWithArray q = new QueueWithArray();
        Person person = new Person();
        Scanner kybd = new Scanner(System.in);
        System.out.print("Join (j), leave (l) or end (e)? ");
        String action = kybd.nextLine();
        while (!action.equalsIgnoreCase("e"))
        {
            if (action.equalsIgnoreCase("j"))
            {
                System.out.print("Enter your name: ");
                String name = kybd.nextLine();
                person.setName(name);
                q.add(name);
                System.out.println("What is your age? ");
                int age = kybd.nextInt();
                person.setAge(age);
                kybd.nextLine();
                System.out.println(name + " is going to the cinema and is " + age + " years old.");
            } 
            else if (action.equalsIgnoreCase("l"))
            {
                if (!q.isEmpty())
                {
                    if (person.getAge() >= 15 ) 
                    {
                        System.out.println(q.remove() + " has left the queue and entered the cinema");
                    }
                    else
                    {
                        System.out.println(q.remove() + " has left the queue, but is not old enough to watch the film");
                    }
                } 
                else
                {
                    System.out.println("Queue empty");
                }
            } 
            else
            {
                System.out.println("Invalid operation");
            }
            System.out.print("Join (j), leave (l) or end (e)? ");
            action = kybd.nextLine();
        }
    }
}

Answer 1

您应该拥有第二个列表，以存储人的年龄。

想象一下您有一个Pidgeonhole，您可以在其中放一张纸上的纸。与指定的年龄有关。

当涉及到每个人时，您每次都会阅读同一张纸。取而代

之类的东西

System.out.print("Enter your name: ");
String name = kybd.nextLine();
System.out.println("What is your age? ");
int age = kybd.nextInt();
qName.add(name);
qAge.add(age);
kybd.nextLine();

或一种更好的方法：

System.out.print("Enter your name: ");
String name = kybd.nextLine();
System.out.println("What is your age? ");
int age = kybd.nextInt();
Person person = new Person(name, age);
qPerson.add(person);
kybd.nextLine();