How to map a String JPA property to a JSON column using Hibernate

Imagine having a tool that can automatically detect JPA and Hibernate performance issues. Wouldn’t that be just awesome?

Well, Hypersistence Optimizer is that tool! And it works with Spring Boot, Spring Framework, Jakarta EE, Java EE, Quarkus, or Play Framework.

So, enjoy spending your time on the things you love rather than fixing performance issues in your production system on a Saturday night!

Introduction

In this article, I want to show you how you can map a String JPA property to a JSON database column using the hibernate-types open-source project.

Although it’s probably more common to use a JsonNode or POJO (Plain Old Java Object) on the Java side, the hibernate-types framework is very flexible and allows you to use a String JPA property type to represent a JSON structure.

Domain Model

Considering we have a book database table that defines a properties column of the jsonb PostgreSQL type.

Book table

The associated Book JPA entity can be mapped as follows:

@Entity(name = "Book")
@Table(name = "book")
@TypeDef(
    name = "json", 
    typeClass = JsonType.class
)
public class Book {

    @Id
    @GeneratedValue
    private Long id;

    @NaturalId
    private String isbn;

    @Type(type = "json")
    @Column(columnDefinition = "jsonb")
    private String properties;

    //Getters and setters omitted for brevity
}

The isbn property is mapped using the Hibernate-specific @NaturalId annotation which allows you to retrieve the entity by its natural identifier.

The properties JPA attribute encodes various book-related properties in a JSON String object. From the JPA @Column definition, we can see that the associated database column is of the type jsonb.

Now, since Hibernate does not provide a native Type to handle JSON database columns, we need to need to use the JsonType offered by the hibernate-types library.

To use the hibernate-types library in your project, just add the following Maven dependency:

<dependency>
    <groupId>com.vladmihalcea</groupId>
    <artifactId>hibernate-types-55</artifactId>
    <version>${hibernate-types.version}</version>
</dependency>

If you’re using an older version of Hibernate, go to the hibernate-types GitHub repository and find the matching hibernate-types dependency for your current Hibernate version.

Testing time

When persisting the following Book entity:

entityManager.persist(
    new Book()
        .setIsbn("978-9730228236")
        .setProperties(
            "{" +
            "   \"title\": \"High-Performance Java Persistence\"," +
            "   \"author\": \"Vlad Mihalcea\"," +
            "   \"publisher\": \"Amazon\"," +
            "   \"price\": 44.99" +
            "}"
        )
);

Hibernate generates the following SQL INSERT statement:

INSERT INTO book (
    isbn, 
    properties, 
    id
) 
VALUES (
    '978-9730228236', 
    '{"title":"High-Performance Java Persistence","author":"Vlad Mihalcea","publisher":"Amazon","price":44.99}', 
    1
)

Notice the Fluent-style API used when creating the Book entity. For more details about building entities using a Fluent-style API, check out this article.

Now, when fetching the previously persisted Book entity:

Book book = entityManager
    .unwrap(Session.class)
    .bySimpleNaturalId(Book.class)
    .load("978-9730228236");

assertTrue(book.getProperties().contains("\"price\": 44.99"));

We can see that the properties attribute is properly populated by the JsonType.

Cool, right?

If you enjoyed this article, I bet you are going to love my Book and Video Courses as well.

Conclusion

Although creating a custom Hibernate Type is straightforward, it’s much more convenient to use the hibernate-types open-source project since you only need to add one dependency and specify which custom Type you want to use via the @TypeDef annotation.

Transactions and Concurrency Control eBook

Leave a Reply

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.