how to serialize class?

问题:

When I insert a List into mongodb, there is a problem:

Exception in thread "main" java.lang.IllegalArgumentException: can't serialize class mongodb.Person
    at org.bson.BasicBSONEncoder._putObjectField(BasicBSONEncoder.java:234)
    at org.bson.BasicBSONEncoder.putIterable(BasicBSONEncoder.java:259)
    at org.bson.BasicBSONEncoder._putObjectField(BasicBSONEncoder.java:198)
    at org.bson.BasicBSONEncoder.putObject(BasicBSONEncoder.java:140)
    at org.bson.BasicBSONEncoder.putObject(BasicBSONEncoder.java:86)
    at com.mongodb.DefaultDBEncoder.writeObject(DefaultDBEncoder.java:27)
    at com.mongodb.OutMessage.putObject(OutMessage.java:142)
    at com.mongodb.DBApiLayer$MyCollection.insert(DBApiLayer.java:252)
    at com.mongodb.DBApiLayer$MyCollection.insert(DBApiLayer.java:211)
    at com.mongodb.DBCollection.insert(DBCollection.java:57)
    at com.mongodb.DBCollection.insert(DBCollection.java:87)
    at com.mongodb.DBCollection.save(DBCollection.java:716)
    at com.mongodb.DBCollection.save(DBCollection.java:691)
    at mongodb.MongoDB.main(MongoDB.java:45)

the class Person is defined as follows:

class Person{
    private String name;
    public Person(String name){
        this.name = name;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
}

The program is :

        DBCollection coll = db.getCollection("test");
        DBObject record = new BasicDBObject();
        List<Person> persons= new ArrayList<Person>();
        persons.add(new Person("Jack"));
        record.put("person", persons);
        coll.save(record);

I can't find the answer from google, so please help me.

回答1:

Just implement Serializable interface in Person class.

Also it will be good to define a serialVersionUID in your class.

AFAIK, while creating POJO class in java, the class should be serializable, if it is going to be transfered over some stream, have a default constructor, and allows access to properties/fields using getter and setter methods.

You might be interested in reading this: Discover the secrets of the Java Serialization API

回答2:

I got the same exception while working with mongodb. I tried making the problematic class serializable but that didn't fix my problem.

Following is what worked for me. Extend the class to be a child of BasicDBObject . Of course this works only if the problem is caused by MongoDB.

extends BasicDBObject 

Original source

http://techidiocy.com/cant-serialize-class-mongodb-illegal-argument-exception/#comment-1298

回答3:

class Person should implement java.io.Serializable interface.

class Person implements Serializable

回答4:

Your Person class definition needs to have implements Serializable in order for it to be serialized, e.g.:

class Person implements Serializable {
   //Rest here
}

Here are some useful links on Java object serialization: Link, Link.

回答5:

The problem here not in "implements Serializable". The problem is that object not converted in the DBObject.

Possible solution:

import com.fasterxml.jackson.databind.ObjectMapper;
import com.mongodb.*;

....

ObjectMapper mapper = new ObjectMapper();
DBObject dboJack = mapper.convertValue(new Person("Jack"), BasicDBObject.class);
...

回答6:

You can achieve this by using following code:

import com.google.gson.annotations.Expose;
import com.mongodb.ReflectionDBObject;

class PersonList extends ReflectionDBObject {
    // person property
    @Expose public java.util.List<Person> person;
}

Now in your mongodb code, you can serialise a Person list as follows

....
PersonList personList = new PersonList();
personList.person = new ArrayList<>();
// add persons to the list
....
....
record.put("personsList", personList);
....
// rest of your code

回答7:

Here is the code example to make Employee object serialized:

public class Employee implements Serializable {

    private int empId;
    private String name;

    public int getEmpId() {
        return empId;
    }

    public String getName() {
        return name;
    }

    public void setEmpId(int empId) {
        this.empId = empId;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Override
    public String toString() {
        return "EMployee id : " + empId + "  \nEmployee Name : " + name;
    }
}

//Another Main Class
public class Main{
    public static void main(String[] args) 
        throws FileNotFoundException, IOException, ClassNotFoundException {

        String filename = "data.txt";
        Employee e = new Employee();
        e.setEmpId(101);
        e.setName("Yasir Shabbir");

        FileOutputStream fos = null;
        ObjectOutputStream out = null;

        fos = new FileOutputStream(filename);
        out = new ObjectOutputStream(fos);
        out.writeObject(e);

        out.close();

        // Now to read the object from file
        // save the object to file
        FileInputStream fis = null;
        ObjectInputStream in = null;

        fis = new FileInputStream(filename);
        in = new ObjectInputStream(fis);
        e = (Employee) in.readObject();
        in.close();

        System.out.println(e.toString());
    }
}

回答8:

First of all you should know why you make class Serializable? Whenever you want to move obeject on network to a file, database, network, process or any other system. In java simple Implementation. Just Implement Serializable interface.