使用@MapsId持久化@OneToOne子实体会在Hibernate中抛出“错误：传递给持久化的分离实体”

提问者：小点点

使用@MapsId持久化@OneToOne子实体会在Hibernate中抛出“错误：传递给持久化的分离实体”

我读了https://vladmihalcea.com/the-best-way-to-map-a-onetoone-relationship-with-jpa-and-hibernate/.我尝试了建议配置（使用Spring数据JPA，hibernate 5.0作为供应商）：

public class PaperSubjectType{
    @Id
    private Long id;

    @OneToOne(fetch = FetchType.LAZY)
    @MapsId
    private PaperSetting paperSetting;
..
}

class PaperSetting{
  @Id
  @GeneratedValue
  private Long id;
..
}

首先我尝试了这个例子：

PaperSetting paperSettingInDb = paperSettingRepository.findOne(1);
PaperSubjectType paperSubjectType = new PaperSubjectType();
paperSubjectType.setSubjectCode("91");
paperSubjectType.setPaperSetting(paperSettingInDb);

paperSubjectTypeRepository.save(paperSubjectType);

错误：传递给持久化的分离实体：纸质设置。级联时，似乎Hibernate将纸质设置视为分离

2如果我想同时创建PaperSubjectType和PaperSsettings，我需要这样做吗：

PaperSetting paperSetting = new PaperSetting();
paperSetting.setxx;
PaperSetting  paperSettingInDbNew = paperSettingRepository.save(paperSetting);
PaperSubjectType paperSubjectType = new PaperSubjectType();
paperSubjectType.setPaperSetting(paperSettingInDbNew);
paperSubjectTypeRepository.save(paperSubjectType);

或者我应该在这种情况下使用双向？谢谢！

共3个答案

匿名用户

我想你可能忘记了将逻辑包装在@Transactional块中

@Transactional
PaperSetting paperSettingInDb = paperSettingRepository.findOne(1);
PaperSubjectType paperSubjectType = new PaperSubjectType();
paperSubjectType.setSubjectCode("91");
paperSubjectType.setPaperSetting(paperSettingInDb);

paperSubjectTypeRepository.save(paperSubjectType);

如果没有crudReposity. findOne（）将打开它自己的短期事务，因此当您获得findOne（）的返回时，实体已经分离，因此出现错误

匿名用户

我尝试了Hibernate 5.2，它的工作原理就像一个魅力。

假设您有这些实体：

@Entity(name = "Person")
public static class Person  {

    @Id
    @GeneratedValue
    private Long id;

    @NaturalId
    private String registrationNumber;

    public Person() {}

    public Person(String registrationNumber) {
        this.registrationNumber = registrationNumber;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getRegistrationNumber() {
        return registrationNumber;
    }
}

@Entity(name = "PersonDetails")
public static class PersonDetails  {

    @Id
    private Long id;

    private String nickName;

    @OneToOne
    @MapsId
    private Person person;

    public String getNickName() {
        return nickName;
    }

    public void setNickName(String nickName) {
        this.nickName = nickName;
    }

    public Person getPerson() {
        return person;
    }

    public void setPerson(Person person) {
        this.person = person;
    }
}

而这个数据访问逻辑：

Person _person = doInJPA( this::entityManagerFactory, entityManager -> {
    Person person = new Person( "ABC-123" );
    entityManager.persist( person );

    return person;
} );

doInJPA( this::entityManagerFactory, entityManager -> {
    Person person = entityManager.find( Person.class, _person.getId() );

    PersonDetails personDetails = new PersonDetails();
    personDetails.setNickName( "John Doe" );
    personDetails.setPerson( person );

    entityManager.persist( personDetails );
} );

测试在HibernateORM中通过得很好。

也许是5.0中的bug得到了修复，所以你最好升级。

匿名用户

1）添加级联选项：

@OneToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
@MapsId
private PaperSetting paperSetting;

2）有了它，您可以在重新创建两个实体时仅保存PaperSubjectType：

PaperSetting paperSetting = new PaperSetting();
paperSetting.setxx;
PaperSubjectType paperSubjectType = new PaperSubjectType();
paperSubjectType.setPaperSetting(paperSettingInDbNew);
paperSubjectTypeRepository.save(paperSubjectType);