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. Hypersistence Optimizer is that tool!


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")
    name = "jsonb", 
    typeClass = JsonBinaryType.class
public class Book {

    private Long id;

    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:


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:

    new Book()
            "{" +
            "   \"title\": \"High-Performance Java Persistence\"," +
            "   \"author\": \"Vlad Mihalcea\"," +
            "   \"publisher\": \"Amazon\"," +
            "   \"price\": 44.99" +

Hibernate generates the following SQL INSERT statement:

    '{"title":"High-Performance Java Persistence","author":"Vlad Mihalcea","publisher":"Amazon","price":44.99}', 

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

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

  1. When this annotation is used with Kotlin:

    @TypeDef(name = "jsonb", typeClass = JsonBinaryType.class)

    It is giving a compile error:

    An annotation argument must be a compile-time constant
    • I’ve never used Kotlin and the Hibernate Types framework offers no guarantee that it will wok on Kotlin. It’s developed and tested only on Java.

    • Use the following mapping:

      @TypeDef(name = "jsonb", typeClass = JsonBinaryType::class)

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