通过消息队列进行序列化和反序列化

Ani*_*n B 3 java serialization rabbitmq deserialization spring-boot

我有一个Employee类如下:

package com.mypackage.rabbitmq.model
import java.io.Serializable;

import javax.xml.bind.annotation.XmlRootElement;


@XmlRootElement
public class Employee implements Serializable{

    /**
    * 
    */
    private static final long serialVersionUID = -2736911235490297622L;
    private int EmpNo;
    private String FirstName;
    private String LastName;
    private int age;
    private String gender;
    private String skill;
    private long phone;
    private String email;
    private double salary;
    //getters and setters
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我在rabbit MQ中发布了员工名单如下:

package com.mypackage.rabbitmq.client.publisher;

//imports

public class Publisher {

    public static void main(String[] args) throws IOException {
        ConnectionFactory factory = new ConnectionFactory();

        Connection con = factory.newConnection("localhost");
        Channel channel = con.createChannel();

        Gson gson  = new GsonBuilder().create();
        Employee employee = null;

        List<Employee> empList = new ArrayList<>();
        ByteArrayOutputStream bos = new ByteArrayOutputStream();
        ObjectOutputStream oos = new ObjectOutputStream(bos);

        String queueName = "TestQueue";
        for(int i=1; i<=10; i++){
            employee = newEmp(i);

            String message =gson.toJson(employee);
            System.out.println("queueName: "+queueName);
            empList.add(employee);

        }
        oos.writeObject(empList);
        channel.basicPublish(1, "", queueName, null, bos.toByteArray());
        System.out.println("[X], sent '"+empList+"'");

        channel.close(0, queueName);
        con.close(0, queueName);
    }

    public static Employee newEmp(int i){

        //logic here
    }
}
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从一个单独的 Spring Boot 应用程序中,我尝试使用员工列表。在消费者应用程序中,我有相同的员工类结构。但是包不一样。

package org.springboot.consumer.model;

import java.io.Serializable;

public class Employee implements Serializable{
    //fields and getters-setters here
}
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简而言之,消费者代码如下:

public class SDPRabbitMQConsumer implements MessageListener {

    @Override
    public void onMessage(Message message) {
        Gson gson  = new GsonBuilder().create();
        try {
            ByteArrayOutputStream os = new ByteArrayOutputStream();
            os.write(message.getBody(), 0, message.getBody().length);
            ByteArrayInputStream is = new ByteArrayInputStream(os.toByteArray());
            ObjectInputStream objInputStream = new ObjectInputStream(is);
            System.out.println("objInputStream.readObject().toString()"+objInputStream.readObject().toString());
            Employee[] employeeArray = gson.fromJson(objInputStream.readObject().toString(), Employee[].class);
            List<Employee> employeeList = new ArrayList<Employee>(Arrays.asList(employeeArray));
            for(Employee employee: employeeList){
                System.out.println(employee);
            }
        } catch (Exception e) {
            e.printStackTrace();
        }

    }

}
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但我收到以下异常:

java.lang.ClassNotFoundException: com.mypackage.rabbitmq.model.Employee
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似乎这是一个序列化和反序列化问题。

我的问题是:

  1. 如果我发布 Employee 而不是 List 我什至不需要序列化类。那么为什么我需要在 List 的情况下进行序列化?
  2. 为什么这个异常?我们是否需要在两端为序列化类维护相同的包结构?
  3. 在消费者端我们需要有serialVersionUID吗?如果是,是否应该与出版方的一致?

提前致谢。

Uli*_*ses 5

1) 虽然 java.util.List 不是 Serializable,但所有标准实现都是,因此您应该能够序列化/反序列化 ArrayList 例如。

2)是的,你需要在同一个包中的同一个类,而且..

3)两端的类必须有相同的serialVersionUID