如何在javaspringboot(jpa-hibernate)中保存(分离的)对象的关系

Question

我想用特定的元素保存两个或多个人之间的关系。一个有技能的人和一个特定地址的人应该相互联系。通常，我会创建一个表来保存每个元素的ID，并在表中创建一行(在带有PHP的普通MySQL中(
如何在Java Spring Boot(JPA Hibernate MySQL(中解决此问题？

当我创建(或者更好地要求(一个"；对象"；(分离的(每个元素的存储库，并希望将其保存在一个新的"存储库"中；储存库"；(数据库(然后我得到了一个错误。

PartnerConnectionServiceImplementation(.java(

@Service
public class PartnerConnectionServiceImpl implements PartnerConnectionService {
@Autowired
private PartnerConnectionRepository partnerConnectionRepository;
@Autowired
private DanceSkillServiceImpl danceSkillDatabaseService;
@Autowired
private AddressLocationServiceImpl addressLocationDatabaseService;
@Autowired
private UserProfileServiceImpl userProfileDatabaseService;
@Override
public Optional<PartnerConnection> connectPartnersWithDanceSkillAndAddressLocation(long userIdPartner1, long danceSkillIdPartner1, long addressLocationIdPartner1, long userIdPartner2, long danceSkillIdPartner2, long addressLocationIdPartner2) {
Optional<UserProfile> userProfile1 = this.userProfileDatabaseService.getUserById(userIdPartner1);
Optional<UserProfile> userProfile2 = this.userProfileDatabaseService.getUserById(userIdPartner2);
Optional<DanceSkill> danceSkill1 = this.danceSkillDatabaseService.getDanceSkillById(danceSkillIdPartner1);
Optional<DanceSkill> danceSkill2 = this.danceSkillDatabaseService.getDanceSkillById(danceSkillIdPartner2);
Optional<AddressLocation> addressLocation1 = this.addressLocationDatabaseService.getAddressLocationById(addressLocationIdPartner1);
Optional<AddressLocation> addressLocation2 = this.addressLocationDatabaseService.getAddressLocationById(addressLocationIdPartner2);
if (
(userProfile1.isPresent()) && (userProfile2.isPresent())
){
Optional<PartnerConnection> theConnection = getPartnerConnectionOfPartners(
userProfile1.get(),
userProfile2.get());
if (theConnection.isPresent()) {
return theConnection;
}
}

if (
(userProfile1.isPresent()) && (userProfile2.isPresent()) &&
(danceSkill1.isPresent()) && (danceSkill2.isPresent()) &&
(addressLocation1.isPresent()) && (addressLocation2.isPresent())
) {
PartnerConnection newPartnerConnection = new PartnerConnection(
null,
userProfile1.get(),
danceSkill1.get(),
addressLocation1.get(),
userProfile2.get(),
danceSkill2.get(),
addressLocation2.get()
);
this.partnerConnectionRepository.save(newPartnerConnection);

return Optional.of(newPartnerConnection);
}
return Optional.empty();
}
...

PartnerConnection(.java(

// indicates the connecitons between partners/ users
@NoArgsConstructor
@AllArgsConstructor
@Data
@Entity
@Table(name = "partner_connection")
public class PartnerConnection {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;

@OneToOne(
fetch = FetchType.LAZY,
cascade = CascadeType.ALL
)
@JoinColumn(
name = "firstmessage_fk",    // foreign key
nullable = true
)
private UserMessage firstMessage;
@ManyToOne(
fetch = FetchType.LAZY,
cascade = CascadeType.ALL
)
@JoinColumn(name = "onepartner_fk",     // foreign key
nullable = false)
private UserProfile firstPartner;
@OneToOne(
fetch = FetchType.LAZY,
cascade = CascadeType.ALL
)
@JoinColumn(
name = "firstpartnerdanceskill_fk",    // foreign key
nullable = false
)
private DanceSkill firstPartnerDanceSkill;
@OneToOne(fetch = FetchType.LAZY,
cascade = CascadeType.ALL)
@JoinColumn(name = "firstpartneraddresslocation_fk",    // foreign key
nullable = false)
private AddressLocation firstPartnerAddressLocation;
@ManyToOne(fetch = FetchType.LAZY,
cascade = CascadeType.ALL)
@JoinColumn(name = "secondpartner_fk",   // foreign key
nullable = false)
private UserProfile secondPartner;
@OneToOne(fetch = FetchType.LAZY,
cascade = CascadeType.ALL)
@JoinColumn(name = "secondpartnerdanceskill_fk",    // foreign key
nullable = false)
private DanceSkill secondPartnerDanceSkill;
@OneToOne(fetch = FetchType.LAZY,
cascade = CascadeType.ALL)
@JoinColumn(name = "secondpartneraddresslocation_fk",    // foreign key
nullable = false)
private AddressLocation secondPartnerAddressLocation;

public PartnerConnection(UserMessage firstMessage, UserProfile firstPartner, DanceSkill firstPartnerDanceSkill, AddressLocation firstPartnerAddressLocation, UserProfile secondPartner, DanceSkill secondPartnerDanceSkill, AddressLocation secondPartnerAddressLocation) {
this.firstMessage = firstMessage;
this.firstPartner = firstPartner;
this.firstPartnerDanceSkill = firstPartnerDanceSkill;
this.firstPartnerAddressLocation = firstPartnerAddressLocation;
this.secondPartner = secondPartner;
this.secondPartnerDanceSkill = secondPartnerDanceSkill;
this.secondPartnerAddressLocation = secondPartnerAddressLocation;
}
}

错误org.springframework.dao.InvalidDataAccessApiUsageException: detached entity passed to persist: ... nested exception is org.hibernate.PersistentObjectException: detached entity passed to persist: ...出现在this.partnerConnectionRepository.save(newPartnerConnection);上

你有什么易于理解的建议吗？

Answer 1

我认为您的方法应该包含@Transactional注释。你已经将所有关系标记为LAZY，所以如果你想获得它们，你需要一个事务将它们从分离的状态加载到托管状态，然后能够将它附加到你想要保存的对象