Json Object Deserialization

Question

I want to desalinize following JSON object to Java Object (Location Class) :

{
  "city": "TEST",
  "country": "TEST",
  "latitude": "1",
  "longitude": "1",
  "name": "TEST",
  "postalCode": "362001",
  "street": "TEST",
  "organization": 3
}

My Java Class is :

public class Location{
@ManyToOne
@ForeignKey(name = "FK_location__organization")
private Organization organization;

@NotNull
@Column(nullable = false)
private String name;

private boolean disabled;

private String street;
private String postalCode;
private String city;
private String country;

private Double latitude;
private Double longitude;

@Override
public String toString()
{
    return name;
}

public static List<Location> findAllLocationsOrderedByName()
{
    return entityManager().createQuery("SELECT o FROM Location o ORDER BY name ASC", Location.class).getResultList();}}

And, Organization Class is :

public class Organization {
String name
}

I got following error when trying to deserialize :

flexjson.JsonNumber cannot be cast to java.util.Map
flexjson.factories.BeanObjectFactory.instantiate(BeanObjectFactory.java:17)
flexjson.ObjectBinder.bind(ObjectBinder.java:95)
flexjson.ObjectBinder.bindIntoObject(ObjectBinder.java:149)
flexjson.ObjectBinder.bind(ObjectBinder.java:95)
flexjson.ObjectBinder.bind(ObjectBinder.java:74)
flexjson.JSONDeserializer.deserialize(JSONDeserializer.java:158)
flexjson.factories.ClassLocatorObjectFactory.instantiate(ClassLocatorObjectFactory.java:38)

I use following code (using Flex JSON) for deserializing :

import flexjson.JSONDeserializer;
import flexjson.JSONSerializer;

public static Location Location.fromJsonToLocation(String json) {
    System.out.println(json);
    return new JSONDeserializer<Location>()
    .use(null, Location.class).deserialize(json);
}

I want to Deserialize that Json Object in to Java Class.. when i tried i got following error flexjson.JsonNumber cannot be cast to java.util.Map flexjson.factories.BeanObjectFactory.instantiate(BeanObjectFactory.java:17) flexjson.ObjectBinder.bind(ObjectBinder.java:95) flexjson.ObjectBinder.bindIntoObject(ObjectBinder.java:149) flexjson.factories.ClassLocatorObjectFactory.instantiate(ClassLocatorObjectFactory.java:38) flexjson.ObjectBinder.bind(ObjectBinder.java:95) flexjson.ObjectBinder.bind(ObjectBinder.java:74) flexjson.JSONDeserializer.deserialize(JSONDeserializer.java:158) — Devloper, Dec 31 '14 at 12:42
Please update the question with this error. Also please add what tool are you using to parse JSON. — mvg, Dec 31 '14 at 12:44
Don't you think it should have been relevant to show us what you did not just what you want? In the question not in cumbersome and hard to read comments would be much better. — Edwin Dalorzo, Dec 31 '14 at 12:44
Please add the code you have used to parse JSON in the question. — mvg, Dec 31 '14 at 12:46
public static Location Location.fromJsonToLocation(String json) { System.out.println(json); return new JSONDeserializer() .use(null, Location.class).deserialize(json); } — Devloper, Dec 31 '14 at 12:49

score 1 · Accepted Answer · answered Dec 31 '14 at 13:10

1

Take a look at organization value in Json. It's integer value but in Java object is Organization

answered Dec 31 '14 at 13:10

Linh

1,024
2
16
28

So,what i need to pass there? Should i pass this json object { "city": "TEST", "country": "TEST", "latitude": "1", "longitude": "1", "name": "TEST", "postalCode": "362001", "street": "TEST", "organization": {"id" : 3 , "name" : "TEST"} } – Devloper Dec 31 '14 at 13:33

Prasad Khode · Answer 2 · 2014-12-31T12:57:31.323

0

You can use ObjectMapper to convert your json data to corresponding Object by using the following snippet:

ObjectMapper mapper = new ObjectMapper();
Location location= mapper.readValue(jsonData, Location.class);

edited Dec 31 '14 at 12:57

answered Dec 31 '14 at 12:51

Prasad Khode

6,602
11
44
59

@Devloper do you see any exception or any error message?? – Prasad Khode Dec 31 '14 at 12:56

Json Object Deserialization

2 Answers2