How to Insert a Record with a Custom ID Using Spring Data JDBC?

How can I insert a record with a custom ID in Spring Data JDBC?

// Example entity class
@Table("products")
public class Product {
    @Id
    private Long id;
    private String name;
    private BigDecimal price;

    // Getters and Setters
}

Spring Data JDBC is a simplified alternative to JPA for database access in Spring applications. Inserting a record with a custom ID involves creating an entity class and using a repository to save instances of that class. This approach provides a lightweight, straightforward means of interacting with databases.

// Example repository interface
import org.springframework.data.repository.CrudRepository;

public interface ProductRepository extends CrudRepository<Product, Long> {
}

// Example of saving a product with a custom ID
Product product = new Product();
product.setId(1001L); // Setting custom ID
product.setName("Laptop");
product.setPrice(new BigDecimal("1200.00"));
productRepository.save(product); // Saves product with custom ID

Causes

• Not setting the ID field before saving the entity.
• Using incorrect annotations that affect ID generation.
• Forgetting to use the @Id annotation on the custom ID field.

Solutions

• Define your entity class with the custom ID field using the @Id annotation.
• Use a repository interface that extends CrudRepository or a similar interface.
• Ensure that you explicitly set your custom ID before saving the entity.

Mistake: Not initializing the ID before saving the object.

Solution: Always set a unique ID before calling the save method.

Mistake: Using an incorrect data type for the ID field.

Solution: Ensure the ID data type matches the type used in the database schema.

Mistake: Omitting the @Id annotation on the ID field.

Solution: Declare the field with the @Id annotation to let Spring Data JDBC know it’s the identity for the entity.