CODEWITHSUNDEEP
javajpa
javajaba
javajaba

Reputation: 85

How to format JPA entity as JSON with database columns?

What is the best way convert a JPA entity to JSON that does not include the related objects but the database columns.

Consider:

public class Question implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Basic(optional = false)
    private Integer id;

    @Basic(optional = false)
    @NotNull
    @Size(min = 1, max = 255)
    @Column(name = "question_title")
    private String questionTitle;

    @Size(max = 500)
    @Column(name = "question_desc")
    private String questionDesc;

    @JoinColumn(name = "requires_question_id", referencedColumnName = "id")
    @ManyToOne
    private Question requiredQuestion;

    ...
}

Converting a Question directly to json may look like this:

{
    "id": 1,
    "questionTitle": "Q1",
    "questionDesc": "A Question 123"
    "requiredQuestion": {
        "id": 2, 
        "questionTitle": "Q2",
        "questionDesc": "A Question 456"
     } 
}

However a more desirable result may look like this:

{
    "id": 1,
    "question_title": "Q1",
    "question_desc": "A Question 123"
    "required_question_id": 2
}

What is the best way to obtain the desirable result?

Create a hashmap with String keys like "question_title", etc and manually assign the object properties, then convert the map to json?

Upvotes: 1

Views: 4865

Answers (3)

erv-Z
erv-Z

Reputation: 87

In this type of application, you should not use the same object for comunicating with the client and for persisting data on your database. The reason is that is possible that your table estructure change for a requirement or for whatever and you have to change your JPA Entity. Also you have to change the client's code because the JPA Entity class change its structure.

You should create DTO, that is a simple POJO used for transfer information (serialization) to client and getting information from the client (deserializing). And use a library like Jackson or gson for transform Java object to Json format and the inverse.

Upvotes: 1

Ruby Kannan
Ruby Kannan

Reputation: 251

Assigning values manually to an object is not a good idea, because you can get your desirable result by projection in hibernate and then you can convert the map to json

return sessionFactory.getCurrentSession()
.createCriteria(Question.class)
.createAlias("requiredQuestion","requiredQuestion")
.setProjection(Projections.projectionList()
    .add(Projections.property("id").as("id"))
    .add(Projections.property("questionTitle").as("questionTitle"))
    .add(Projections.property("questionDesc").as("questionDesc"))
    .add(Projections.property("requiredQuestion.id").as("requiredQuestionid")))
    .setResultTransformer(Criteria.ALIAS_TO_ENTITY_MAP)
.list();

This is the code to get the result what you are expecting and you can convert it to json.

Upvotes: 0

Predrag Maric
Predrag Maric

Reputation: 24423

You can use Jackson JSON Processor to serialize your entities to JSON just like any other POJO. To have full control on the output format, write a custom serializer for Question class. It would look something like this 

public class QuestionSerializer extends JsonSerializer<Question> {
    @Override
    public void serialize(Question value, JsonGenerator jgen, SerializerProvider provider) 
      throws IOException, JsonProcessingException {
        jgen.writeStartObject();
        jgen.writeNumberField("id", value.getId());
        jgen.writeStringField("question_title", value.getQuestionTitle());
        jgen.writeStringField("question_desc", value.getQuestionDesc());
        jgen.writeNumberField("required_question_id", value.getRequiredQuestion().getId());
        jgen.writeEndObject();
    }
}

Upvotes: 2

Related Questions