I am trying to work with @EmbeddedId, this is my code as follows,
create table TBL_EMPLOYEE_002(
ID integer generated always as identity (start with 100,increment by 10),
COUNTRY varchar(50),
NAME varchar(50),
constraint PK_EMP_00240 primary key(ID,COUNTRY)
)
The Embedded class as follows,
@Embeddable
public class EmployeeIdTwo implements Serializable{
public EmployeeIdTwo(){}
public EmployeeIdTwo(String country){
this.empCountry = country;
}
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="ID")
private Integer employeeId;
@Column(name="COUNTRY",length=50)
private String empCountry;
// implementation of hashCode and equals and only getters
...
}
employee Entity as follows,
@Entity
@Table(name="TBL_EMPLOYEE_002")
public class EmployeeEntitySix implements Serializable{
public EmployeeEntitySix(){}
public EmployeeEntitySix(EmployeeIdTwo id,String name){
this.id = id;
this.employeeName = name;
}
@EmbeddedId
private EmployeeIdTwo id;
@Column(name="NAME")
private String employeeName;
// getters and setters
}
this is the code written in main method,
private static void storVal(EntityManager em){
EmployeeEntitySix employee = new EmployeeEntitySix(new EmployeeIdTwo("KENYA"), "Henry Olaanga");
em.persist(employee);
}
but once i run an above code i get an exception as follows,
Caused by: javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: Attempt to modify an identity column 'ID'.
Can you please let me know where i am getting wrong, if my EmbeddedId class contain an autogenerated column, than what should be the approach.
Just to know I am using hibernate as persistence provider and JPA as persistence API
