如何解决Hibernate" ERROR o.h.e.jdbc.spi.SqlExceptionHelper - 重复条目"例外

时间:2017-04-10 15:02:24

标签: mysql jdbc spring-data-jpa hibernate-mapping sqlexception

现在已经2天了,我测试解决了这个问题,但没有任何效果。 我将这些实体置于双向关系中。

@Entity
@Table( name = "T_user" )
public class T_user implements Serializable {
   @Id
   @GeneratedValue( strategy = GenerationType.IDENTITY )
   private Long               id_user;

   @OneToOne( targetEntity = T_infosProfile.class, mappedBy = "user", fetch = FetchType.LAZY, orphanRemoval = true )
   private T_infosProfile     infosProfile;

   public void setInfosProfile( T_infosProfile infosProfile ) {
    this.infosProfile = infosProfile;
   }
}

@Entity
@Table( name = "T_infosProfile" )
public class T_infosProfile {

   @Id
   @GeneratedValue( strategy = GenerationType.IDENTITY )
   private Long       id_infos;

   @OneToOne( targetEntity = T_user.class, fetch = FetchType.LAZY )
   @JoinColumn( name = "id_user", unique = true, nullable = false )
   private T_user user;

     @Column( name = "pourcentage_completion", length = 3, updatable = true, unique = false )
     private Integer    pourcentageCompletion;

     @Column( name = "infos_identite", nullable = false, updatable = true, length = 1, columnDefinition = "TINYINT(1)" )
     private Boolean    infosIdentiteCompleted;

     @Column( name = "infos_coordonnees", nullable = false, updatable = true, length = 1, columnDefinition = "TINYINT(1)" )
     private Boolean    infosCoordonneesCompleted;

     @Column( name = "infos_bancaires", nullable = false, updatable = true, length = 1, columnDefinition = "TINYINT(1)" )
     private Boolean    infosBancaireCompleted;

     @Column( name = "infos_chicowa", nullable = false, updatable = true, length = 1, columnDefinition = "TINYINT(1)" )
     private Boolean    infosCompleted;

   //the setter method
   public void setUser( T_user user) {
      this.user = user;
      this.infosIdentiteCompleted = true;
      this.pourcentageCompletion = 0;
      this.infosCoordonneesCompleted = this.user.getInfosCoordonnees() == null ? false : true;
      this.infosBancaireCompleted = this.user.getInfosBancaire() == null ? false : true;
      this.infosCompleted = this.user.getInfosCompleted() == null ? false : true;

      if ( this.infosCoordonneesCompleted == true ) {
        this.pourcentageCompletion = 50;
    }
      if ( this.infosBancaireCompleted == true && this.pourcentageCompletion == 50 ) {
        this.pourcentageCompletion = 75;
    }
      if ( this.infosChicowaCompleted == true && this.pourcentageCompletion == 75 ) {
        this.pourcentageCompletion = 100;
    }
  }
}

我也有这种方法用于更新用户的infosProfile:

@Service
public class UIServiceImpl implements UIService {

  @Autowired
  private TinfosProfileRepository   infosProfileRepository;

  @Override
  public T_infosProfile saveInfosPorfile( T_user connectedUser ){
    T_infosProfile infosProfile = connectedUser.getInfosProfile();
    infosProfile.setUser( connectedUser );
    connectedUser.setInfosProfile( infosProfile );
    try {
        return infosProfileRepository.save( infosProfile );
    } catch ( Exception e ) {
        throw new ExceptionsDAO( e.getMessage() );
    }
}

但是当我调用这个方法时,它适用于一种情况,但对于另一种情况,我得到了这个例外:

10-04-2017 16:27:43 [http-nio-8080-exec-3] WARN  o.h.e.jdbc.spi.SqlExceptionHelper - SQL Error: 1062, SQLState: 23000
10-04-2017 16:27:43 [http-nio-8080-exec-3] ERROR  o.h.e.jdbc.spi.SqlExceptionHelper - Duplicate entry '8' for key 'UK_bkariaocmnn2rlh477hvoqkyf'

我只想更新当前用户的infosProfile属性。 如果只更新了一个infosProfile属性,但是当我同时更新2个属性并尝试saveInfosProfile()时,会引发异常。

请帮忙。

3 个答案:

答案 0 :(得分:1)

查看错误消息,并且它自我解释,因为它表示该列中已存在值8。由于您在该特定列上定义了UNIQUE CONSTRAINT

,因此会抛出该错误
Duplicate entry '8' for key 'UK_bkariaocmnn2rlh477hvoqkyf'

附加:查看您的约束名称UK_bkariaocmnn2rlh477hvoqkyf。这就是当您在创建约束时没有命名约束时会发生什么。

答案 1 :(得分:0)

好吧,在失去4天之后,我找到了一个似乎有效的解决方案。在这里我的解决方案:

@Override
public T_infosProfile saveInfosPorfile( T_user connectedUser ){
  T_infosProfile infosProfile = infosProfileRepository.findByUser(connectedUser); // I had to load by the user before saving it.
  infosProfile.setUser( connectedUser );
  connectedUser.setInfosProfile( infosProfile );
  try {
    return infosProfileRepository.save( infosProfile );
  } catch ( Exception e ) {
    throw new ExceptionsDAO( e.getMessage() );
  }
}

对我而言,这是一种奇怪的行为,因为用户与保存在表格中的用户相同。谢谢Hibernate!

答案 2 :(得分:0)

在我的春季项目中争论了一段时间后,我能够使用下面的代码来解决我面临的问题。

from celery import task

@task(name='automate')
def automate():
    automate_all()
    run_second_task()