Question

在ManyToMany关联期间，我在Hibernate方面遇到了一些麻烦：我想让一个用户与其联系人联系。关联表用于具有关联的creation_date及其状态（例如，活动，非活动等）

休眠版本：5.2.17
春季靴：2.0.5

我的班级用户：

@Entity(name = "user")
public class User implements Serializable {

@ManyToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
@JoinTable(name = "assoc_user_user", joinColumns = {@JoinColumn(name = 
"id.myself.id")}, inverseJoinColumns = {@JoinColumn(name = 
"id.contact.id")})
private List<AssocUserUser> contacts;

@ManyToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
@JoinTable(name = "assoc_user_user", joinColumns = {@JoinColumn(name = 
"id.contact.id")}, inverseJoinColumns = {@JoinColumn(name = 
"id.myself.id")})
private List<AssocUserUser> contactOf;

}

我的关联类AssocUserUser：

@Entity(name = "assoc_user_user")
public class AssocUserUser implements Serializable {

private static final long serialVersionUID = 1L;

@EmbeddedId
private AssocUserUserId id;

@Column(name = "creation_date", nullable = false)
private LocalDateTime creationDate;

@Column(name = "status")
@Enumerated(EnumType.STRING)
private ContactStatusEnum status;

}

我的EmbeddedId类AssocUserUserId：

@Embeddable
public class AssocUserUserId implements Serializable {

private static final long serialVersionUID = 1L;

@ManyToOne
@JoinColumn(name = "user_id_myself", nullable = false)
private User myself;

@ManyToOne
@JoinColumn(name = "user_id_contact", nullable = false)
private User contact;

}

我的错误：

由于：org.hibernate.AnnotationException：外键引用 com .....用户的com ..... AssocUserUser用户的错误编号柱。应该是2

Answer 1

好吧，这个问题已经在Spring-Data-JPA ManyToMany relationship with extra column上提出并回答了，但我想它与自我参考有所不同。我不是所有JoinTable和JoinColumn批注的忠实拥护者，只是因为它们通常是多余的，恕我直言，恕我直言，它意在更改默认值。我的答案不包括那些注释，因此请根据需要添加它们。请注意，对于Spring-Data-Jpa，您需要每个实体的存储库，因此您需要关联实体的存储库。这与JPA中的单个实体管理器不同。这会稍微影响您处理关联实体的方式。在这里，我们通过创建和保留新的AssocUserUser实体来保留新的关系。

我完成课程的方式是：

@Entity
public class User {
    @Id @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    @OneToMany(mappedBy="myself")
    private Set<AssocUserUser> contacts;

@Entity
public class AssocUserUser {
    @EmbeddedId
    private AssocUserUserId id = new AssocUserUserId();
    @ManyToOne @MapsId("myselfId")
    private User myself;
    @ManyToOne @MapsId("contactId")
    private User contact;

@SuppressWarnings("serial")
@Embeddable
public class AssocUserUserId implements Serializable {
    private Long myselfId;
    private Long contactId;

两个实体的

和Repositories

public interface UserRepository extends JpaRepository<User, Long> {
    @Query("select u from User u left outer join fetch u.contacts where u.id = :userId")
    User getContactsForUser(@Param("userId") Long userId);

public interface AssocUserUserRepository extends JpaRepository<AssocUserUser, AssocUserUserId> {

如何创建关联

private void update() {
    User user1 = new User();
    User contact1 = new User();
    userRepo.save(user1);
    userRepo.save(contact1);
    AssocUserUser assoc = new AssocUserUser();
    assoc.setMyself(user1);
    assoc.setContact(contact1);
    assocUserUserRepo.save(assoc);
}

如何以Unidirectional方式阅读关联

private void read() {
    User me = new User();
    me.setId(1L);
    AssocUserUser assoc = new AssocUserUser();
    assoc.setMyself(me);
    List<AssocUserUser> contacts = assocUserUserRepo.findAll(Example.of(assoc));
    System.out.println(contacts.size());
}

并以Bidirectional的方式阅读关联：

private void readBi() {
    User me = userRepo.getContactsForUser(1L);
    System.out.println(me.getContacts().size());
}

与往常一样，查看日志：

create table assoc_user_user (contact_id bigint not null, myself_id bigint not null, primary key (contact_id, myself_id))
create table user (id bigint generated by default as identity, primary key (id))
alter table assoc_user_user add constraint FKaccetv956cu63fwiejjfrm0mi foreign key (contact_id) references user
alter table assoc_user_user add constraint FK1absxfuktrjnom8vwtjfqx5l0 foreign key (myself_id) references user

insert into user (id) values (null)
insert into user (id) values (null)
select assocuseru0_.contact_id as contact_1_0_0_, assocuseru0_.myself_id as myself_i2_0_0_ from assoc_user_user assocuseru0_ where assocuseru0_.contact_id=? and assocuseru0_.myself_id=?
select user0_.id as id1_1_0_ from user user0_ where user0_.id=?
select user0_.id as id1_1_0_ from user user0_ where user0_.id=?
insert into assoc_user_user (contact_id, myself_id) values (?, ?)

select assocuseru0_.contact_id as contact_1_0_, assocuseru0_.myself_id as myself_i2_0_ from assoc_user_user assocuseru0_ inner join user user1_ on assocuseru0_.myself_id=user1_.id where user1_.id=1
select user0_.id as id1_1_0_ from user user0_ where user0_.id=?
select user0_.id as id1_1_0_ from user user0_ where user0_.id=?

select user0_.id as id1_1_0_, contacts1_.contact_id as contact_1_0_1_, contacts1_.myself_id as myself_i2_0_1_, contacts1_.myself_id as myself_i2_0_0__, contacts1_.contact_id as contact_1_0_0__ from user user0_ left outer join assoc_user_user contacts1_ on user0_.id=contacts1_.myself_id where user0_.id=?
select user0_.id as id1_1_0_ from user user0_ where user0_.id=?

带有EmbeddedId的AssociationTable可以实现ManyToMany-Spring-Data-Jpa

1 个答案: