在对象添加到队列后访问对象变量

Question

我想知道，当Customer对象名称和食物已经添加到队列中时，我将如何获取它？那么，假设我想在第一个客户对象添加到队列后使用from name和food元素打印一个字符串？队列peek方法是占位符，因为我不知道在将对象添加到队列后如何访问对象的名称和食物。如果我打印peek方法，它只会给我内存位置，而不是对象的食物或名称。

结果是这样的：

"你想加工什么：披萨还是沙拉？

沙拉

詹姆斯的沙拉做好了！"

代码：

主要类别：

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();
        }
    }
}

客户类别：

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LinkedList.peek()，它确实返回了正确的对象。我相信你只是看到了对象的散列，因为你打印了Customer对象，它没有重新定义.toString()：你使用的是Object.toString()，它返回你看到的散列。

如果你总是想把你的Customer表示为它的名字和食物选择，那么按照Zack Macomber的建议，在Customer中重新定义.toString()，或者选择System.out.println(queue.peek().getName() + " choosed " + queue.peek().getFood())或类似的东西。