CODEWITHSUNDEEP
jpajava-persistence-api
Mrye
Mrye

Reputation: 727

JPA : Entity extend with entity

How can I extend an entity with another entity but both of them referring to the same table ? Is it possible ? The structure is something like this :

@Entity
@Table(name = "users")
@NamedQuery(name="User.findAll", query="SELECT u FROM User u")
public class User implements Serializable{
    private int id;
    private String name;
}

@Entity
@Table(name = "users")
@NamedQuery(name="SubUser.findAll", query="SELECT su FROM SubUser su")
public class SubUser extends User {

    @Override
    @Id  
    @GeneratedValue(strategy=GenerationType.AUTO)
    public int getId() {
      return super.getId();
    }

    //- Other fields and getter setter

}

I tried this way Extend JPA entity to add attributes and logic

but I got this exception 

org.hibernate.mapping.SingleTableSubclass cannot be cast to org.hibernate.mapping.RootClass

Update 1

I already put the @Id for the SubUser because the @Entity shows this exception

The entity has no primary key attribute defined

Upvotes: 1

Views: 7442

Answers (2)

Neil McGuigan
Neil McGuigan

Reputation: 48256

  • Add the @Inheritance annotation to the super class
  • Implement Serializable
  • Add a getter for id (you don't need a setter necessarily)
  • id should be Integer, not int, so that you can represent unassigned ids with null.

Code:

@Entity
@Table(name = "users")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class User implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id  
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Integer id;

    private String name;

    public Integer getId() {
      return id;
    }

    public String getName() {
      return name;
    }

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


@Entity
public class SubUser extends User {

}

Upvotes: 2

Neil Stockton
Neil Stockton

Reputation: 11531

Any basic JPA docs would describe inheritance, discriminators and use of @Id.

@Entity
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="DISCRIM", discriminatorType=DiscriminatorType.STRING)
@DiscriminatorValue("User")
@Table(name="users")
@NamedQuery(name="User.findAll", query="SELECT u FROM User u")
public class User implements Serializable{
    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private int id;

    private String name;
}

@Entity
@DiscriminatorValue("SubUser")
@NamedQuery(name="SubUser.findAll", query="SELECT su FROM SubUser su")
public class SubUser extends User {

}

Upvotes: -1

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