java - 有没有办法发送可序列化类的不同子类的实例？

Question

我已经构建了一个客户端-服务器应用程序，具有有效的连接设置和处理功能。我希望客户端发送不同类型的请求，每个请求都有唯一的字段，但所有这些都继承自“Request”父类(super class)。有办法做到这一点吗？

例如，我有一个实现可序列化的 Foo 类

class Foo implements Serializable{

public String name;

public Foo() {
    this.name = "Default";
}

public Foo(String name) {
    this.name = name;
}
}

以及扩展它的两个子类

class BarOne extends Foo {

public String name;
public int id;

public Bar(String name, int id) {
    super(name);
    this.id = id;
}
}

class BarTwo extends Foo {

public String name;
public String lastName;

public Bar(String name, String lastName) {
    super(name);
    this.lastName = lastName;
}
}

有没有一种方法可以让客户端通过 ObjectOutputStream 发送 BarOne 和 BarTwo 的实例，并让服务器确定它接收到的是其中一个的实例还是另一个的实例？

Answer 1

这里没有限制，你可以毫无困难地序列化子类。

在这里，我提供了一个快速示例，使用与您相同的结构(但有具体示例)。

抽象类。

public abstract class Pojo implements Serializable {

    private static final long serialVersionUID = -4947411931465651278L;

    protected int id;

    public Pojo() {
        // TODO Auto-generated constructor stub
    }

    public Pojo(int id) {
        this.id = id;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }
}

第一个子类Person

class Person extends Pojo {

    private static final long serialVersionUID = -7814628079202659483L;

    private String name;
    private int age;

    public Person(int id, String name, int age) {
        super(id);
        this.name = name;
        this.age = age;
    }

    @Override
    public String toString() {
        return "Person [name=" + name + ", age=" + age + ", id=" + id + "]";
    }
}

第二个子类地址

class Address extends Pojo {

    private static final long serialVersionUID = -8266402026827561883L;

    private String address;
    private String city;

    public Address(int id, String address, String city) {
        super(id);
        this.address = address;
        this.city = city;
    }

    @Override
    public String toString() {
        return "Address [address=" + address + ", city=" + city + ", id=" + id
                + "]";
    }
}

然后，您可以为此示例创建一个集合来保存这两种类型的实例:

List<Pojo> pojos = new ArrayList<>();
pojos.add(new Address(1, "Address 1", "city1"));
pojos.add(new Address(2, "Address 2", "city2"));
pojos.add(new Person(1, "Name1", 5));
pojos.add(new Person(2, "Name2", 10));
pojos.add(new Person(3, "Name3", 15));

您可以使用文件进行序列化(因为我没有时间创建套接字系统;))使用:

private static void serialize(String filename, Serializable data)
        throws IOException {
    try (OutputStream outStream = new FileOutputStream(filename);
            ObjectOutputStream fileObjectOut = new ObjectOutputStream(
                    outStream)) {
        fileObjectOut.writeObject(data);
    }
}

private static Object deserialize(String filename) throws IOException,
        ClassNotFoundException {
    try (InputStream inStream = new FileInputStream(filename);
            ObjectInputStream fileObjectIn = new ObjectInputStream(inStream)) {
        return fileObjectIn.readObject();
    }
}

让列表序列化和反序列化以查看类型是否已恢复:

serialize("data.ser", pojos);
List<Pojo> tmp = deserialize("data.ser");
System.out.println(tmp);

您将看到具有正确类型的实例的结果:

[
    Address [address=Address 1, city=city1, id=1], 
    Address [address=Address 2, city=city2, id=2], 
    Person [name=Name1, age=5, id=1], Person [name=Name2, age=10, id=2], 
    Person [name=Name3, age=15, id=3]
]

所以你可以对任何东西做同样的事情，只序列化一个Person，你需要首先恢复Object并使用instanceof查看恢复的类型。

Pojo p1 = new Person(1, "Name1", 10);
serialize("person.ser", p1);
Object o1 = deserialize("person.ser");

System.out.println(o1);
System.out.println(o1.getClass());
System.out.println(o1 instanceof Person);

输出

Person [name=Name1, age=10, id=1]
class serial.Person
true