When working with Hibernate and Jakarta Persistence API (JPA), you might encounter the error: Could not determine recommended JdbcType for Class. This error typically occurs when Hibernate is unable to map a Java type to a corresponding JDBC type. In this article, we will explore how to resolve this issue.
1. Problem Statement
We have an Orders entity class that contains a list of Product objects. The Product class is not an entity, and we want to store this list in JSON format in a MySQL database.
Here is the Order entity:
@Entity
public class Orders implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Long id;
@Column(columnDefinition = "json")
private Map<String, Product> products;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public Map<String, Product> getProducts() {
return products;
}
public void setProducts(Map<String, Product> products) {
this.products = products;
}
}
And the Product class:
public class Product {
String productName;
String category;
int quantity;
BigDecimal price;
// Getters and setters
}
And here is the SQL Table for Orders
CREATE TABLE `Orders` (
id BIGINT AUTO_INCREMENT PRIMARY KEY,
products JSON
);
Here is our main class:
public class HibernateJpaExample {
public static void main(String[] args) {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("jpaexample-unit");
EntityManager em = emf.createEntityManager();
em.getTransaction().begin();
Product product1 = new Product();
product1.setProductName("Laptop");
product1.setCategory("Electronics");
product1.setQuantity(10);
product1.setPrice(new BigDecimal("999.99"));
Product product2 = new Product();
product2.setProductName("Mouse");
product2.setCategory("Accessories");
product2.setQuantity(50);
product2.setPrice(new BigDecimal("19.99"));
Map<String, Product> firstOrder = new HashMap<>();
firstOrder.put("product1", product1);
Map<String, Product> secondOrder = new HashMap<>();
secondOrder.put("product2", product2);
Orders order = new Orders();
order.setProducts(firstOrder);
Orders order1 = new Orders();
order1.setProducts(secondOrder);
em.persist(order);
em.persist(order1);
em.getTransaction().commit();
em.close();
emf.close();
}
}
When you run the application, you encounter the following exception:
Exception in thread "main" org.hibernate.type.descriptor.java.spi.JdbcTypeRecommendationException: Could not determine recommended JdbcType for Java type 'java.util.Map<java.lang.String, com.jcg.Product>' at org.hibernate.type.descriptor.java.spi.UnknownBasicJavaType.getRecommendedJdbcType(UnknownBasicJavaType.java:50)
2. Solution
To resolve this issue, we can use the @JdbcTypeCode annotation from Hibernate to specify that the products field should be stored as a JSON column. This involves configuring Hibernate to recognize the JSON type and converting the product list to JSON format when persisting and retrieving data.
We would also need to add the Jackson Library to the project for JSON Column Handling. When using Hibernate Types to handle JSON columns in the database, Jackson is used internally for converting Java objects to JSON strings and JSON strings to Java objects. This ensures that the @JdbcTypeCode annotation with SqlTypes.JSON works seamlessly.
First, ensure you have the necessary dependencies in the pom.xml file:
<dependency>
<groupId>org.hibernate.orm</groupId>
<artifactId>hibernate-core</artifactId>
<version>6.5.2.Final</version>
</dependency>
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.17.2</version>
</dependency>
Update Entity Class:
Next, Use the @JdbcTypeCode annotation to specify the JSON type for the products field.
@Entity
public class Orders implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Long id;
@Column(columnDefinition = "JSON")
@JdbcTypeCode(SqlTypes.JSON)
private Map<String, Product> products;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public Map getProducts() {
return products;
}
public void setProducts(Map products) {
this.products = products;
}
}
In this Orders class, the products field is annotated with @Column(columnDefinition = "JSON") and @JdbcTypeCode(SqlTypes.JSON). This setup instructs Hibernate to handle the products field as a JSON column in the database.
When the Orders entity is persisted, Hibernate will automatically convert the list of Product objects into a JSON string and store it in the products column of the database table. Conversely, when an Orders entity is retrieved from the database, Hibernate will convert the JSON string back into a list of Product objects.
2.1 Expected Output
During the application’s execution, Hibernate will output the following SQL statements to the console if the hibernate.show_sql property has been set to true in the persistence.xml file. You will see SQL commands such as:
Hibernate: insert into Orders (products) values (cast(? as json)) Hibernate: insert into Orders (products) values (cast(? as json)) Hibernate: update Orders set products=cast(? as json) where id=? Hibernate: update Orders set products=cast(? as json) where id=?
These statements indicate that the Orders entity is being inserted into the database, with the products field being stored as a JSON string.
In the MySQL database, the Orders table will have new rows representing the inserted Orders entity. The products column will contain a JSON string representing the list of Product objects. For example, if we run the following SQL statement:
SELECT * FROM `Order`;
This query returns:
This output shows that the Orders entity was successfully persisted with its products field correctly stored as a JSON string in the database.
3. Conclusion
In this article, we explored the “Could Not Determine Recommended JdbcType for Class” error in JPA. Through an example, we demonstrated how to reproduce this error and provided a solution to resolve it. Understanding and addressing these issues ensures our JPA and Hibernate applications function smoothly, reducing runtime errors and improving data integrity.
4. Download the Source Code
This article addresses the ‘JPA could not determine recommended JdbcType for class’ error.
You can download the full source code of this example here: jpa could not determine recommended jdbctype for class
