JPA Model for Document Structure Entity

Introduction

JPA Model for Document Structure Entity.

Before software can be reusable it first has to be usable.
-- Ralph Johnson

In SQL, sometimes you want to store directly JSON documents without creating a relational table (like MongoDB or key-value pair). JSON documents support embedded fields, so related data and lists of data can be stored with the document instead of an external table.

In this snippet, you'll find how to map a JSON document structure model in JPA without having a direct relational table.

Step 1: Define Entity

Document structure of employee entity that has enum, collection, map, nested object, and nested collection.

Step 2: Model entity for JPA

enum EmployeeStatus {
    ACTIVE, FIRED
}

@Embeddable
@Access(AccessType.FIELD)
@NoArgsConstructor(access = AccessLevel.PROTECTED)
@Getter
@AllArgsConstructor
@EqualsAndHashCode
@ToString
class Head {
    private String name;
    @Embedded
    private Code code;
}

@Embeddable
@Access(AccessType.FIELD)
@NoArgsConstructor(access = AccessLevel.PROTECTED)
@Getter
@AllArgsConstructor
@EqualsAndHashCode
@ToString
class Department {
    private String name;
    private Head head;
    @ElementCollection(fetch = FetchType.EAGER)
    private Set<Address> addresses;
    @Embedded
    private Code code;
}

@Embeddable
@Access(AccessType.FIELD)
@NoArgsConstructor(access = AccessLevel.PROTECTED)
@Getter
@AllArgsConstructor
@EqualsAndHashCode
@ToString
class Code {
    private String code;
}

@Embeddable
@Access(AccessType.FIELD)
@NoArgsConstructor(access = AccessLevel.PROTECTED)
@Getter
@AllArgsConstructor
@EqualsAndHashCode
@ToString
class Address {
    private String addressLine;
    @Embedded
    private Code code;
}

@Entity
@Table(name = "EMPLOYEES")
@Access(AccessType.FIELD)
@Getter
@NoArgsConstructor(access = AccessLevel.PROTECTED)
@EqualsAndHashCode(of = "id")
@ToString(of = "id")
class Employee {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    private String name;
    @Embedded
    private Code code;
    @Embedded
    private Department department;
    @ElementCollection(fetch = FetchType.EAGER)
    private Map<String, Integer> skills;
    @ElementCollection(fetch = FetchType.EAGER)
    private Set<Address> addresses;
    @Enumerated(EnumType.STRING)
    private EmployeeStatus status;
    private LocalDate joiningDate;
    public Employee(String name,
                    Department department,
                    Map<String, Integer> skills,
                    Set<Address> addresses,
                    Code code) {
        this.name = name;
        this.department = department;
        this.skills = skills;
        this.addresses = addresses;
        this.joiningDate = LocalDate.now();
        this.status = EmployeeStatus.ACTIVE;
        this.code = code;
    }
}

Step 3: JPA Repository

interface EmployeeRepository extends JpaRepository<Employee, Long> {
}

Step 4: Application Properties

To support JPA Multiple Embedded fields with a prefix without having @AttributeOverride annotations:

spring.jpa.hibernate.naming.implicit-strategy=org.hibernate.boot.model.naming.ImplicitNamingStrategyComponentPathImpl

Step 4: Usage

@Spring BootApplication
@EnableJpaRepositories(considerNestedRepositories = true)
@Slf4j
public class JPAModelForDocumentStructureEntity {

    private final EmployeeRepository employeeRepository;
    
    public JPAModelForDocumentStructureEntity(EmployeeRepository employeeRepository) {
        this.employeeRepository = employeeRepository;
    }
    
    @EventListener
    public void run(ApplicationReadyEvent readyEvent) {
        Map<String, Integer> skills = new HashMap<>();
        skills.put("Java", 90);
        skills.put("Python", 80);
        Address addressLine1 = new Address("addressLine1", new Code("1-CODE"));
        Address addressLine2 = new Address("addressLine2", new Code("2-CODE"));
        Set<Address> addresses1 = Set.of(addressLine1, addressLine2);
        Set<Address> addresses2 = Set.of(addressLine1, addressLine2);
        Head head = new Head("h-name", new Code("H-CODE"));
        Department department = new Department("d-name", head, addresses1, new Code("D-CODE"));
        Employee employee = new Employee("e-name", department, skills, addresses2, new Code("E-CODE"));
        // Save
        Employee saved = employeeRepository.save(employee);
        // Find by Id
        Employee findById = employeeRepository.findById(saved.getId()).get();
        log(findById);
    }
    
    private void log(Employee employee) {
        log.info("{}", employee);
        log.info("-----------------");
        log.info("Name: {}", employee.getName());
        log.info("Code: {}", employee.getCode());
        log.info("Department: {}", employee.getDepartment());
        log.info("Skills: {}", employee.getSkills());
        log.info("Addresses: {}", employee.getAddresses());
        log.info("Status: {}", employee.getStatus());
        log.info("Joining Date: {}", employee.getJoiningDate());
        log.info("-----------------");
    }
}

Findings

Use Set instead of List to fix the Hibernate MultipleBagFetchException
Use field-based access @Access(AccessType.FIELD) strategy why ?
- Better readability of your code
- Omit getter or setter methods
- No need to mark utility methods as @Transient