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How to deserialize a list using GSON or another JSON library in Java?

Posted by: admin November 30, 2017 Leave a comment

Questions:

I can serialize a List in my servlet on GAE, but I can’t deserialize it. What am I doing wrong?

This is my class Video in GAE, which is serialized:

package legiontube;

import java.util.Date;

import javax.jdo.annotations.IdGeneratorStrategy;
import javax.jdo.annotations.IdentityType;
import javax.jdo.annotations.PersistenceCapable;
import javax.jdo.annotations.Persistent;
import javax.jdo.annotations.PrimaryKey;

@PersistenceCapable(identityType = IdentityType.APPLICATION)
public class Video {

    @PrimaryKey
    private String id;

    @Persistent
    private String titulo;

    @Persistent
    private String descricao;

    @Persistent
    private Date date;

 /**
  * @param id
  * @param titulo
  * @param descricao
  * @param date
  */

    public Video(){};

 public Video(String id, String titulo, String descricao, Date date) {
  //super();
  this.id = id;
  this.titulo = titulo;
  this.descricao = descricao;
  this.date = date;
 }

 public String getId() {
  return id;
 }

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

 public String getTitulo() {
  return titulo;
 }

 public void setTitulo(String titulo) {
  this.titulo = titulo;
 }

 public String getDescricao() {
  return descricao;
 }

 public void setDescricao(String descricao) {
  this.descricao = descricao;
 }

 public Date getDate() {
  return date;
 }

 public void setDate(Date date) {
  this.date = date;
 }

}

This is my class Video in my other application, where I try to deserialize:

package classes;

import java.util.Date;

public class Video {
 private String id;
 private String titulo;
 private String descricao;
 private Date date;

 /**
  * @param id
  * @param titulo
  * @param descricao
  * @param date
  */
 public Video(String id, String titulo, String descricao, Date date) {
  //super();
  this.id = id;
  this.titulo = titulo;
  this.descricao = descricao;
  this.date = date;
 }

 public String getId() {
  return id;
 }
 public void setId(String id) {
  this.id = id;
 }
 public String getTitulo() {
  return titulo;
 }
 public void setTitulo(String titulo) {
  this.titulo = titulo;
 }
 public String getDescricao() {
  return descricao;
 }
 public void setDescricao(String descricao) {
  this.descricao = descricao;
 }
 public Date getDate() {
  return date;
 }
 public void setDate(Date date) {
  this.date = date;
 }

}
Answers:

With Gson, you’d just need to do something like:

List<Video> videos = gson.fromJson(json, new TypeToken<List<Video>>(){}.getType());

You might also need to provide a no-arg constructor on the Video class you’re deserializing to.

Questions:
Answers:

Another way is to use an array as a type, e.g.:

Video[] videoArray = gson.fromJson(json, Video[].class);

This way you avoid all the hassle with the Type object, and if you really need a list you can always convert the array to a list, e.g.:

List<Video> videoList = Arrays.asList(videoArray);

IMHO this is much more readable.

Questions:
Answers:

Try this one-liner

List<Video> videos = Arrays.asList(new Gson().fromJson(json, Video[].class));

Reference:

  1. Method Arrays#asList
  2. Constructor Gson
  3. Method Gson#fromJson (source json may be of type JsonElement, Reader, or String)
  4. Interface List
  5. JLS – Arrays
  6. JLS – Generic Interfaces
Questions:
Answers:

Be careful using the answer provide by @DevNG.
Arrays.asList() returns internal implementation of ArrayList that doesn’t provide some useful methods like add(), delete(), etc.
In order to get real ArrayList instance you need to write something like this:

List<Video> = new ArrayList<>(Arrays.asList(videoArray));