Question

我有以下问题

@Entity
@Table(name="tb_pessoa", schema="public")
@Inheritance(strategy = InheritanceType.JOINED)
public class Pessoa implements Serializable {
   private static final long serialVersionUID = 1L;

   @Id
   @SequenceGenerator(name = "tb_pessoa_id_seq", sequenceName = "tb_pessoa_id_seq", schema="public")
   @GeneratedValue(strategy = GenerationType.AUTO, generator = "tb_pessoa_id_seq")
   @Column(name = "id", nullable = false)
   private Integer id;

   ...

@Entity  
@Table(name="tb_pessoafisica", schema="public")
@PessoaFisicaAnnotation
@PrimaryKeyJoinColumn(name = "id")
public class PessoaFisica extends Pessoa implements Serializable {
   private static final long serialVersionUID = 1L;

   @Column(name = "email")
   private String email;

   ...

假设 Pessoa 表的记录 id 1到500.当我运行以下代码时：

PessoaFisica pessoaFisica = new PessoaFisica();
pessoaFisica.setEmail("teste@123.com");

...

em.persist(pessoa);

发生了以下错误：

17:26:13,613 ERROR [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (http--10.36.1.49-8180-1) ERROR: duplicate key value violates unique constraint "tb_pessoa_pkey"
    Detalhe: Key (id)=(438) already exists.
17:26:13,614 WARN  [com.arjuna.ats.arjuna] (http--10.36.1.49-8180-1) ARJUNA012125: TwoPhaseCoordinator.beforeCompletion - failed for SynchronizationImple< 0:ffff7f000101:-7f6ae4e3:558080ed:4f, org.hibernate.engine.transaction.synchronization.internal.RegisteredSynchronization@331af73a >: javax.persistence.PersistenceException: org.hibernate.exception.ConstraintViolationException: ERROR: duplicate key value violates unique constraint "tb_pessoa_pkey"
  Detalhe: Key (id)=(438) already exists.

为什么尝试使用ID 438 而不是 501 ???

有谁知道可能会发生什么？

非常感谢

CREATE TABLE tb_pessoa
(
    id integer NOT NULL,
    CONSTRAINT tb_pessoa_pkey PRIMARY KEY (id)
)

CREATE TABLE tb_pessoafisica
(
    email character varying(64) NOT NULL,
    CONSTRAINT tb_pessoafisica_pkey PRIMARY KEY (id),
    CONSTRAINT fkee8c050f70415778 FOREIGN KEY (id)
        REFERENCES tb_pessoa (id) MATCH SIMPLE
)

CREATE SEQUENCE tb_pessoa_id_seq
  INCREMENT 1
  MINVALUE 1
  MAXVALUE 9223372036854775807
  START 500
  CACHE 1;

Log Hibernate：

07:50:52,463 INFO  [stdout] (http--10.36.1.49-8180-2) Hibernate: 
07:50:52,463 INFO  [stdout] (http--10.36.1.49-8180-2)     select
07:50:52,464 INFO  [stdout] (http--10.36.1.49-8180-2)         nextval ('public.tb_pessoa_id_seq')
07:50:52,497 INFO  [stdout] (http--10.36.1.49-8180-2) Hibernate: 
07:50:52,498 INFO  [stdout] (http--10.36.1.49-8180-2)     insert 
07:50:52,499 INFO  [stdout] (http--10.36.1.49-8180-2)     into
07:50:52,499 INFO  [stdout] (http--10.36.1.49-8180-2)         public.tb_pessoa
07:50:52,500 INFO  [stdout] (http--10.36.1.49-8180-2)         (id) 
07:50:52,500 INFO  [stdout] (http--10.36.1.49-8180-2)     values
07:50:52,501 INFO  [stdout] (http--10.36.1.49-8180-2)         (?)
07:50:52,519 WARN  [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (http--10.36.1.49-8180-2) SQL Error: 0, SQLState: 23505
07:50:52,521 ERROR [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (http--10.36.1.49-8180-2) ERROR: duplicate key value violates unique constraint "tb_pessoa_pkey"
    Detalhe: Key (id)=(438) already exists.

Answer 1

如果您要保留新实体，请尝试将ID设置为null。

PessoaFisica pessoaFisica = new PessoaFisica();
pessoaFisica.setID(null);
pessoaFisica.setEmail("teste@123.com");

...

em.persist(pessoa);

当休眠尝试持久化时，这将确保对象是新的。

休眠 - 重复键值违反唯一约束

1 个答案: