Question

我试图了解如何获取客户对象名称和食物，当它已添加到队列中时？所以说我想在第一个客户对象的名称和食物元素添加到队列后使用它来打印字符串？队列偷看方法是占位符，因为我不确定在将对象的名称和食物添加到队列后如何访问该对象的名称和食物。如果我打印peek方法，它只是给我内存位置，而不是对象食物或名称。

结果将是这样的：

＆＃34;您想要处理什么：披萨或沙拉？

沙拉

詹姆斯的沙拉已经完成！＆＃34;

代码：

主要课程：

import java.util.Scanner;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.LinkedList;
import java.util.Queue;

public class Main {

    public static void main(String[] args) throws FileNotFoundException {
        File customerTxt = new File("customer.txt");
        Queue<Customer> pizza = new LinkedList<Customer>();
        Queue<Customer> salad = new LinkedList<Customer>();
        try {
            Scanner readCus = new Scanner(customerTxt);
            Scanner readFood = new Scanner(System.in);
            while (readCus.hasNextLine()) {
                String line = readCus.nextLine();
                String[] strArray = line.split(",");
                String customerName = strArray[0];
                String customerFood = strArray[1];
                Customer cus = new Customer(customerName, customerFood);
                if (customerFood.equalsIgnoreCase("salad")) {
                    salad.add(cus);
                }
                if (customerFood.equalsIgnoreCase("pizza")) {
                    pizza.add(cus);
                }
            }
            if (pizza.isEmpty() == false && salad.isEmpty() == false) {
                System.out.println("What kind of food would you like to make?");
                String foodChoice = readFood.nextLine();
                if (foodChoice.equalsIgnoreCase("salad")) {
                    System.out.println(salad.peek());
                }
                if (foodChoice.equalsIgnoreCase("pizza")) {
                    System.out.println(salad.peek());
                }
            }
            if (pizza.isEmpty() == true && salad.isEmpty() == false) {
                System.out.println("There are no Pizzas left to process. I will just finish the rest of the Salads");
                while (salad.isEmpty() == false) {
                    System.out.println(salad.peek());
                }
            }
            if (pizza.isEmpty() == false && salad.isEmpty() == true) {
                System.out.println("There are no Salads left to process. I will just finish the rest of the Pizzas");
                while (pizza.isEmpty() == false) {
                    System.out.println(pizza.peek());
                }
            }
        }

        catch (FileNotFoundException e) {
            e.printStackTrace();
        }
    }
}

客户类：

public class Customer {

    public String name = "";

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String food = "";

    public String getFood() {
        return food;
    }

    public void setFood(String food) {
        this.food = food;
    }

    public Customer(String customerName, String customerFood) {
        this.name = customerName;
        this.food = customerFood;
    }



    }

Answer 1

根据override func touchesBegan(touches: Set<UITouch>, withEvent event: UIEvent?) { for touch in touches { let location = touch.locationInNode(self) let touchedNode = nodeAtPoint(location) touchedNode.removeFromParent() }，它确实返回了正确的对象。我相信你只是看到了对象的哈希，因为你打印了LinkedList.peek()对象，它没有重新定义Customer：你正在使用.toString()来返回你看到的哈希值。 / p>

如果您始终想要将Object.toString()作为其名称和食物选择，或者选择.toString()或者Customer，则可以按照Zack Macomber的建议在Customer重新定义System.out.println(queue.peek().getName() + " choosed " + queue.peek().getFood())类似的东西。

将对象添加到队列后访问对象变量？

1 个答案: