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mysql - 如何解决Hibernate "ERROR o.h.e.jdbc.spi.SqlExceptionHelper - Duplicate entry "异常

转载 作者:行者123 更新时间:2023-11-29 10:53:44 25 4
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我已经尝试解决这个问题两天了,但没有任何效果。我将这些实体隐含在双向关系中。

@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;
}
}

还有

@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;
}
}
}

我还有这个方法,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,我得到了这个异常:

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。它工作正常的情况是,仅更新一个 infosProfile 的属性,但当我同时更新 2 个属性并尝试 saveInfosProfile() 时,会引发异常。

请帮忙。

最佳答案

查看错误消息,它是不言自明的,因为它表示值 8 已存在于该列中。由于您在该特定列上定义了 UNIQUE CONSTRAINT ,因此它会抛出该错误

Duplicate entry '8' for key 'UK_bkariaocmnn2rlh477hvoqkyf'

其他:查看您的约束名称UK_bkariaocmnn2rlh477hvoqkyf。这就是当您在创建约束时未命名约束时会发生的情况。

关于mysql - 如何解决Hibernate "ERROR o.h.e.jdbc.spi.SqlExceptionHelper - Duplicate entry "异常,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43326635/

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