The best way to map a Composite Primary Key with JPA and Hibernate

(Last Updated On: January 1, 2018)

Introduction

One of my readers asked me to help him map a Composite Primary Key using JPA and Hibernate. Because this is a recurrent question, I decided to write a blog post in which I describe this mapping is more detail.

Domain Model

A relational database composite key contains two or more columns which together for the primary key of a given table.

employee_phone_composite_key

In the diagram above, the employee table has a Composite Primary Key, which consists of two columns:

  • department_id
  • employee_number

Every Employee can also have a Phone, which uses the same composite key to reference its owning Employee.

To map this database table mapping, we need to isolate the compound key into an @Embeddable first:

@Embeddable
public class EmployeeId implements Serializable {

    @Column(name = "company_id")
    private Long companyId;

    @Column(name = "employee_number")
    private Long employeeNumber;

    public EmployeeId() {
    }

    public EmployeeId(Long companyId, Long employeeId) {
        this.companyId = companyId;
        this.employeeNumber = employeeId;
    }

    public Long getCompanyId() {
        return companyId;
    }

    public Long getEmployeeNumber() {
        return employeeNumber;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof EmployeeId)) return false;
        EmployeeId that = (EmployeeId) o;
        return Objects.equals(getCompanyId(), that.getCompanyId()) &&
                Objects.equals(getEmployeeNumber(), that.getEmployeeNumber());
    }

    @Override
    public int hashCode() {
        return Objects.hash(getCompanyId(), getEmployeeNumber());
    }
}

The Embeddable must be `Serializable` and we need to provide an implementation for `equals` and `hashCode`.

The Employee mapping looks as follows:

@Entity(name = "Employee")
@Table(name = "employee")
public class Employee {

    @EmbeddedId
    private EmployeeId id;

    private String name;

    public EmployeeId getId() {
        return id;
    }

    public void setId(EmployeeId id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

The @EmbeddedId is used to instruct Hibernate that the Employee entity uses a compound key.

The Phone mapping is rather straightforward as well:

@Entity(name = "Phone")
@Table(name = "phone")
public class Phone {

    @Id
    @Column(name = "`number`")
    private String number;

    @ManyToOne
    @JoinColumns({
        @JoinColumn(
            name = "company_id",
            referencedColumnName = "company_id"),
        @JoinColumn(
            name = "employee_number",
            referencedColumnName = "employee_number")
    })
    private Employee employee;

    public Employee getEmployee() {
        return employee;
    }

    public void setEmployee(Employee employee) {
        this.employee = employee;
    }

    public String getNumber() {
        return number;
    }

    public void setNumber(String number) {
        this.number = number;
    }
}

The Phone uses the number as an entity identifier since every phone number and the @ManyToOne mapping uses the two columns that are part of the compound key.

Testing time

To see how it works, consider the following persistence logic:

doInJPA(entityManager -> {
    Employee employee = new Employee();
    employee.setId(new EmployeeId(1L, 100L));
    employee.setName("Vlad Mihalcea");
    entityManager.persist(employee);
});

doInJPA(entityManager -> {
    Employee employee = entityManager.find(
        Employee.class, new EmployeeId(1L, 100L));
    Phone phone = new Phone();
    phone.setEmployee(employee);
    phone.setNumber("012-345-6789");
    entityManager.persist(phone);
});

doInJPA(entityManager -> {
    Phone phone = entityManager.find(Phone.class, "012-345-6789");
    assertNotNull(phone);
    assertEquals(new EmployeeId(1L, 100L), phone.getEmployee().getId());
});

Which generates the following SQL statements:

INSERT INTO employee (name, company_id, employee_number)
VALUES ('Vlad Mihalcea', 1, 100)

SELECT e.company_id AS company_1_0_0_ ,
       e.employee_number AS employee2_0_0_ ,
       e.name AS name3_0_0_
FROM   employee e
WHERE  e.company_id = 1
       AND e.employee_number = 100

INSERT INTO phone (company_id, employee_number, `number`) 
VALUES (1, 100, '012-345-6789')

SELECT p.number AS number1_1_0_ ,
       p.company_id AS company_2_1_0_ ,
       p.employee_number AS employee3_1_0_ ,
       e.company_id AS company_1_0_1_ ,
       e.employee_number AS employee2_0_1_ ,
       e.name AS name3_0_1_
FROM   phone p
LEFT OUTER JOIN employee e 
ON     p.company_id = e.company_id AND p.employee_number = e.employee_number
WHERE  p.number = '012-345-6789'

Mapping relationships using the Composite Key

We can even map relationships using the information provided within the Composite Key itself. In this particular example, the company_id references a Company entity which looks as follows:

@Entity(name = "Company")
@Table(name = "company")
public class Company {

    @Id
    private Long id;

    private String name;

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Company)) return false;
        Company company = (Company) o;
        return Objects.equals(getName(), company.getName());
    }

    @Override
    public int hashCode() {
        return Objects.hash(getName());
    }
}

We can have the Composite Key mapping referencing the Company entity withing the Employee entity:

@Entity(name = "Employee")
@Table(name = "employee")
public class Employee {

    @EmbeddedId
    private EmployeeId id;

    private String name;

    @ManyToOne
    @JoinColumn(name = "company_id",insertable = false, updatable = false)
    private Company company;

    public EmployeeId getId() {
        return id;
    }

    public void setId(EmployeeId id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

Notice that the @ManyToOne association instructs Hibernate to ignore inserts and updates issued on this mapping since the company_id is controlled by the @EmbeddedId.

Mapping a relationships inside @Embeddable

But that’s not all. We can even move the @ManyToOne inside the @Embeddable itself:

@Embeddable
public class EmployeeId implements Serializable {

    @ManyToOne
    @JoinColumn(name = "company_id")
    private Company company;

    @Column(name = "employee_number")
    private Long employeeNumber;

    public EmployeeId() {
    }

    public EmployeeId(Company company, Long employeeId) {
        this.company = company;
        this.employeeNumber = employeeId;
    }

    public Company getCompany() {
        return company;
    }

    public Long getEmployeeNumber() {
        return employeeNumber;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof EmployeeId)) return false;
        EmployeeId that = (EmployeeId) o;
        return Objects.equals(getCompany(), that.getCompany()) &&
                Objects.equals(getEmployeeNumber(), that.getEmployeeNumber());
    }

    @Override
    public int hashCode() {
        return Objects.hash(getCompany(), getEmployeeNumber());
    }
}

Now, the Employee mapping will no longer require the extra @ManyToOne association since it’s offered by the entity identifier:

@Entity(name = "Employee")
@Table(name = "employee")
public class Employee {

    @EmbeddedId
    private EmployeeId id;

    private String name;

    public EmployeeId getId() {
        return id;
    }

    public void setId(EmployeeId id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

The persistence logic changes as follows:

Company company = doInJPA(entityManager -> {
    Company _company = new Company();
    _company.setId(1L);
    _company.setName("vladmihalcea.com");
    entityManager.persist(_company);
    return _company;
});

doInJPA(entityManager -> {
    Employee employee = new Employee();
    employee.setId(new EmployeeId(company, 100L));
    employee.setName("Vlad Mihalcea");
    entityManager.persist(employee);
});

doInJPA(entityManager -> {
    Employee employee = entityManager.find(
        Employee.class, 
        new EmployeeId(company, 100L)
    );
    Phone phone = new Phone();
    phone.setEmployee(employee);
    phone.setNumber("012-345-6789");
    entityManager.persist(phone);
});

doInJPA(entityManager -> {
    Phone phone = entityManager.find(Phone.class, "012-345-6789");
    assertNotNull(phone);
    assertEquals(new EmployeeId(company, 100L), phone.getEmployee().getId());
});

If you enjoyed this article, I bet you are going to love my book as well.

Conclusion

Knowing how to map a composite key is very important because many database schemas use this type of primary key. As demonstrated by this blog post, such a mapping is not complicated at all.

Code available on GitHub.

Subscribe to our Newsletter

* indicates required
10 000 readers have found this blog worth following!

If you subscribe to my newsletter, you'll get:
  • A free sample of my Video Course about running Integration tests at warp-speed using Docker and tmpfs
  • 3 chapters from my book, High-Performance Java Persistence, 
  • a 10% discount coupon for my book. 
Get the most out of your persistence layer!

Advertisements

28 thoughts on “The best way to map a Composite Primary Key with JPA and Hibernate

  1. Very nice! I was a bit confused about @ManyToOne inside the @Embeddable because of this sentence of the JPA2.2 section 11.1.17 “Relationship mappings defined within an embedded id class are not supported.”. Well, turns out they are supported then 🙂

  2. Thank you for this interesting article.

    I have a question: can a part of the compound key be auto generated ? for example the emloyeeNumber

    I’ve tried this code:

    @Embeddable
    public class EmployeeID implements Serializable{
    
         @Column(name="c_ID")
         private Long companyId;
    
         @GeneratedValue(strategy = GenerationType.IDENTITY)
         @Column(name="e_ID")
         private Long employeeNumber ;
    
         public EmployeeID( Long employeeId) {
            this.employeeNumber = employeeId ;
         }
    }
    
    doInJPA(entityManager -> {
        Employee employee = new Employee();
        employee.setId(new EmployeeId(100L));
        employee.setName("Vlad Mihalcea");
        entityManager.persist(employee);
    })
    

    And, I got this exception:

     ConstraintViolationException: could not execute statement
     MySQLIntegrityConstraintViolationException: Column 'id' cannot be null
    
      1. This works great with mysql but with MSSQL i’m stuck.
        The scenario is same but i’m getting error Hibernate:
        select
        next value for hibernate_sequence
        Sep 03, 2017 4:03:49 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
        WARN: SQL Error: 208, SQLState: S0002
        Sep 03, 2017 4:03:49 AM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
        ERROR: Invalid object name ‘hibernate_sequence’.
        Exception in thread “main” javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: could not extract ResultSet

        My DB has lot of data and its been used from multiple services. I cannot create a custom sequence in db.

        I’ve spend almost 1 week trying to figure out this with sql server.
        Any help would be greatly appreciated.

      2. Looks like you don’t have that sequence in the DB. Of course you need to create it if you want to use the SEQUENCE generator. Otherwise, you can use IDENTITY. Nevertheless, this problem is not strictly related to this article which is about composite identifiers, not auto-generated ones.

      3. Thanks for your reply.
        But I’ve situation like follow
        I’ve a company table in MSSQL server.

        CompayId | CompanyRegNo | CompanyName

        Here CompanyId is autogenerated
        and the composite primary key is CompanyId + CompanyRegNo.

        Now Initially I tried to created an embedded key as

        @Embeddable
        public class CompanyPrimaryKey implements Serializable {

        /**
         * 
         */
        private static final long serialVersionUID = 1L;
        
        
        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        @Column(name = "company_id")
        private Long companyId;
        
        @Column(name = "company_reg_no")
        private Long companyRegNo;
        
        ... 
        Constructors, gettters and setters.
        

        }

        and used this embedded key in other entities
        as

        @Entity(name = “company_master”)
        public class Company implements Serializable {

        /**
         * 
         */
        private static final long serialVersionUID = 34343434343L;
        
        @Column(name = "company_name")
        private String companyName;
        
        @EmbeddedId
        private CompanyPrimaryKey key
        
        ****Rest of the class
        

        }

        When Embedded key consist of a identity column hibernate doesn’t work.

        then I came around a git reference from you blog

        https://github.com/hibernate/hibernate-orm/blob/ceaeb81e3362ff187004ea3479b2afeeba5aa8a6/documentation/src/test/java/org/hibernate/userguide/mapping/identifier/IdClassGeneratedValueTest.java#L77

        Using IdClass it works for mysql.
        But fails in MSSQL.

        How can we have an identity field as a part of embedded key which works for MSSQL?

        Thanks for your valuable articles and the awesome book on persistence performance.
        Learned a lot from that.

      4. Since this is a different topic than what I described in this article, I will answer this question on StackOverflow. Blog comments are not a good way to answer questions in more detail.

        So, you should post the question on StackOverflow and send me a link.

  3. Thanks for a Great Article!

    I have a similar situation, but my Phone entity doesn’t have an Id. It has only the primary foreign composite key. My JPA is complaining Phone has to have an id, In this case can i use a PhoneId or EmployeeId class ? Thanks!

Leave a Reply

Your email address will not be published. Required fields are marked *