CODEWITHSUNDEEP
javahibernate
Alfredo M
Alfredo M

Reputation: 568

how to set a @entity without @id element?

I have this bean:

@Entity
@Table(name = "accesos")
public class Acceso implements Serializable {
    /** */
    @Column(name = "idUser")
    private String idUser;
    /** */
    @ManyToOne
    @JoinColumn(name = "idArea")
    private Area area;
    /** */
    @ManyToOne
    @JoinColumn(name = "idRol")
    private Rol rol;

But I get this error:

Caused by: org.hibernate.AnnotationException: No identifier specified for entity: com...Acceso

How can I set this bean? What I need is based on the user ID get all the ROL-AREA that he has access.

I tried change the @Entity to @Embedded, but when I make the search no result is returned, and even in the log is no SQL sentence executed.

Upvotes: 6

Views: 31315

Answers (3)

MikkolidisMedius
MikkolidisMedius

Reputation: 4844

JPA Specifications state that all Entities must have an identifier (JSR 317, section 2.4). It can be a single column or a composite key.

You can either put an idAcceso identifier in the Acceso entity or not make Acceso an entity but rather a "component" (which is the purpose of the @Embeddable annotation). Components do not require an ID but cannot be queried separately (i.e. you cannot do select a from Acceso a but rather you need to query for User and then use the accessor method user.getAccesos().

You cannot substitute @Entity with @Embedded in this context.

@Embeddable
public class Acceso {
  // ...
}

@Entity
public class User {
  @Id protected String id;
  // ...

  @ElementCollection
  @CollectionTable(
    name="USER_ACCESSES",
    joinColumns=@JoinColumn(name="USER_ID")
  protected Set<Acceso> accesos = new HashSet<Acceso>();
}

Upvotes: 1

slartidan
slartidan

Reputation: 21566

You have to have an identity for each bean, there is no way around. You can however use a combined key, if none of your fields is unique.

If the combination of all your fields is unique, then try to annotate all fields with @Id. Take as few fields as possible, but as many as required to make the combination unique.

Upvotes: 6

slarge
slarge

Reputation: 644

You don't have an id specified and you MUST so add @Id annotation onto idUser 

@Id
@Column(name = "idUser")
private String idUser;

Upvotes: -2

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