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?
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
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>
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();
}
}
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;
}
}
}
最好的方法是使用Apache Commons Lang中的SerializationUtils。
序列化:
byte[] data = SerializationUtils.serialize(yourObject);
反序列化:
YourObject yourObject = SerializationUtils.deserialize(data)
如前所述,这需要Commons Lang库。它可以使用Gradle导入:
compile 'org.apache.commons:commons-lang3:3.5'
Maven:
<!-- https://mvnrepository.com/artifact/org.apache.commons/commons-lang3 -->
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-lang3</artifactId>
<version>3.5</version>
</dependency>
Jar文件
这里提到了更多的方法
或者,可以导入整个集合。参考这个连结
准备要发送的字节数组:
ByteArrayOutputStream bos = new ByteArrayOutputStream();
ObjectOutputStream out = null;
try {
out = new ObjectOutputStream(bos);
out.writeObject(yourObject);
out.flush();
byte[] yourBytes = bos.toByteArray();
...
} finally {
try {
bos.close();
} catch (IOException ex) {
// ignore close exception
}
}
从字节数组创建一个对象:
ByteArrayInputStream bis = new ByteArrayInputStream(yourBytes);
ObjectInput in = null;
try {
in = new ObjectInputStream(bis);
Object o = in.readObject();
...
} finally {
try {
if (in != null) {
in.close();
}
} catch (IOException ex) {
// ignore close exception
}
}
Spring框架org.springframework.util.SerializationUtils
byte[] data = SerializationUtils.serialize(obj);