假设我有一个可序列化的类AppMessage。

我想通过套接字将它作为字节[]传输到另一台机器,在那里它从接收到的字节重建。

我怎么才能做到呢?


当前回答

如果你使用spring,在spring-core中有一个util类可用。你可以简单地

import org.springframework.util.SerializationUtils;

byte[] bytes = SerializationUtils.serialize(anyObject);
Object object = SerializationUtils.deserialize(bytes);

其他回答

如果你使用Java >= 7,你可以使用try和资源改进接受的解决方案:

private byte[] convertToBytes(Object object) throws IOException {
    try (ByteArrayOutputStream bos = new ByteArrayOutputStream();
         ObjectOutputStream out = new ObjectOutputStream(bos)) {
        out.writeObject(object);
        return bos.toByteArray();
    } 
}

反过来说:

private Object convertFromBytes(byte[] bytes) throws IOException, ClassNotFoundException {
    try (ByteArrayInputStream bis = new ByteArrayInputStream(bytes);
         ObjectInputStream in = new ObjectInputStream(bis)) {
        return in.readObject();
    } 
}

我想通过套接字将它作为字节[]传输到另一台机器

// When you connect
ObjectOutputStream oos = new ObjectOutputStream(socket.getOutputStream());
// When you want to send it
oos.writeObject(appMessage);

从接收到的字节重新构建。

// When you connect
ObjectInputStream ois = new ObjectInputStream(socket.getInputStream());
// When you want to receive it
AppMessage appMessage = (AppMessage)ois.readObject();

Spring框架org.springframework.util.SerializationUtils

byte[] data = SerializationUtils.serialize(obj);

Java 8+的代码示例:

public class Person implements Serializable {

private String lastName;
private String firstName;

public Person() {
}

public Person(String firstName, String lastName) {
    this.firstName = firstName;
    this.lastName = lastName;
}

public void setFirstName(String firstName) {
    this.firstName = firstName;
}

public String getFirstName() {
    return firstName;
}

public String getLastName() {
    return lastName;
}

public void setLastName(String lastName) {
    this.lastName = lastName;
}

@Override
public String toString() {
    return "firstName: " + firstName + ", lastName: " + lastName;
}
}


public interface PersonMarshaller {
default Person fromStream(InputStream inputStream) {
    try (ObjectInputStream objectInputStream = new ObjectInputStream(inputStream)) {
        Person person= (Person) objectInputStream.readObject();
        return person;
    } catch (IOException | ClassNotFoundException e) {
        System.err.println(e.getMessage());
        return null;
    }
}

default OutputStream toStream(Person person) {
    try (OutputStream outputStream = new ByteArrayOutputStream()) {
        ObjectOutput objectOutput = new ObjectOutputStream(outputStream);
        objectOutput.writeObject(person);
        objectOutput.flush();
        return outputStream;
    } catch (IOException e) {
        System.err.println(e.getMessage());
        return null;
    }

}

}

另一个有趣的方法来自com.fasterxml.jackson.databind.ObjectMapper

byte[] data = new ObjectMapper().writeValueAsBytes(JAVA_OBJECT_HERE)

Maven的依赖

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
</dependency>