fbrundu fbrundu - 1 year ago 142
JSON Question

Right way to write JSON deserializer in Spring or extend it

I am trying to write a custom JSON deserializer in Spring. I want to use default serializer for most part of fields and use a custom deserializer for few properties. Is it possible?
I am trying this way because, most part of properties are values, so for these I can let Jackson use default deserializer; but few properties are references, so in the custom deserializer I have to query a database for reference name and get reference value from database.

I'll show some code if needed.


Answer Source

I've searched a lot and the best way I've found so far is on this article:

Class to serialize

package net.sghill.example;

import net.sghill.example.UserDeserializer
import net.sghill.example.UserSerializer
import org.codehaus.jackson.map.annotate.JsonDeserialize;
import org.codehaus.jackson.map.annotate.JsonSerialize;

@JsonDeserialize(using = UserDeserializer.class)
public class User {
    private ObjectId id;
    private String   username;
    private String   password;

    public User(ObjectId id, String username, String password) {
        this.id = id;
        this.username = username;
        this.password = password;

    public ObjectId getId()       { return id; }
    public String   getUsername() { return username; }
    public String   getPassword() { return password; }

Deserializer class

package net.sghill.example;

import net.sghill.example.User;
import org.codehaus.jackson.JsonNode;
import org.codehaus.jackson.JsonParser;
import org.codehaus.jackson.ObjectCodec;
import org.codehaus.jackson.map.DeserializationContext;
import org.codehaus.jackson.map.JsonDeserializer;

import java.io.IOException;

public class UserDeserializer extends JsonDeserializer<User> {

    public User deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException {
        ObjectCodec oc = jsonParser.getCodec();
        JsonNode node = oc.readTree(jsonParser);
        return new User(null, node.get("username").getTextValue(), node.get("password").getTextValue());

Edit: Alternatively you can look at this article which uses new versions of com.fasterxml.jackson.databind.JsonDeserializer.