CODEWITHSUNDEEP
javahibernatejpauuid
Luciano Borges
Luciano Borges

Reputation: 827

How to use a UUID field without being a primary key with JPA?

I have an entity that need to have an unique key (not primary) of UUID type.

@Entity
public class MyEntity {

    @Id
    @NotNull
    @GeneratedValue(strategy = SEQUENCE, generator = "seq_entity")
    @SequenceGenerator(name = "seq_entity", sequenceName = "seq_entity", allocationSize = 1)
    private Long id;

    @NotNull
    @Type(type = "pg-uuid")
    @Column(name = "uu_id", unique = true)
    private UUID uuid;

    @NotNull
    @Size(max = 30)
    private String name;

    // gets and sets

}

When I persist this entity how can be seen below in my DAO class:

@Transactional
public class EntityDAO {

    @Inject
    private EntityManager em;

    public void insert(MyEntity myEntity) { //myEntity comes only with name attribute 
        myEntity.setUUID(UUID.randomUUID()); //I'd like to generate automatically by the database
        em.persist(myEntity);
    }

}

Is ocurring the inserting in the database but the following error appears on the console:

09:09:43,529 SEVERE [br.gov.frameworkdemoiselle.exception] (http-/127.0.0.1:8080-1) Erro interno do servidor: org.yaml.snakeyaml.error.YAMLException: No JavaBean properties found in java.util.UUID
    at org.yaml.snakeyaml.introspector.PropertyUtils.getProperties(PropertyUtils.java:97)
    at org.yaml.snakeyaml.introspector.PropertyUtils.getProperties(PropertyUtils.java:87)
    at org.yaml.snakeyaml.representer.Representer.getProperties(Representer.java:243)

Upvotes: 5

Views: 5536

Answers (2)

Takouo Gnegneri Franck Kris-ke
Takouo Gnegneri Franck Kris-ke

Reputation: 137

After several searches, here are some proposals:

Ex1

@Id
@Column(columnDefinition = "BINARY(16)")
private UUID id = UUID.randomUUID();

Ex2

@Id
@GeneratedValue(generator = "uuid")
@GenericGenerator(name = "uuid", strategy = "uuid")
@Column(columnDefinition = "CHAR(32)")
private String id;

Link : enter link description here

Upvotes: -2

Alexander Petrov
Alexander Petrov

Reputation: 9482

You need to Map the UID to String. You need to use a UserType if there is no standard mapping from UID to String.

Or alternativly just use String instead of UUID for the attribute.

You can read the alternatives of mapping UUID in chapter 3.10 of Hibernate documentation

You can use EntityListener instead of setting the ID in the service you can set it in the @PrePersist method.

    @Entity
public class MyEntity {

    @PrePersist
    private assignUIID(){
       myEntity.setUUID(UUID.randomUUID());
    }
}

Upvotes: 3

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