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

(Last Updated On: February 7, 2019)
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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 = "jsonb", 
    typeClass = JsonBinaryType.class
)
public static class Book {

    @Id
    @GeneratedValue
    private Long id;

    @NaturalId
    private String isbn;

    @Type(type = "jsonb")
    @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 JsonBinaryType 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-52</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
)

The JsonBinaryType binds the JSON String using the setObject method of the PreparedStatement object.

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 JsonBinaryType.

Cool, right?

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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.

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6 Comments on “How to map a String JPA property to a JSON column using Hibernate

  1. Pingback: Java Weekly, Issue 267 | Baeldung

  2. Great article.

    I think it should also address the difficulty for searching (like queries based on the json columns), and how this will be RDBMS specific.

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