与同一实体的Hibernate递归多对多关联

Question

另一个休眠问题...：P

使用Hibernate的Annotations框架，我有一个User实体。 每个User可以有一个朋友集合：其他User的集合。 但是，我还无法弄清楚如何在User类中创建一个由User列表组成的多对多关联（使用用户友好中间表）。

这是User类及其注释：

@Entity
@Table(name="tbl_users")
public class User {

    @Id
    @GeneratedValue
    @Column(name="uid")
    private Integer uid;

    ...

    @ManyToMany(
            cascade={CascadeType.PERSIST, CascadeType.MERGE},
            targetEntity=org.beans.User.class
    )
    @JoinTable(
            name="tbl_friends",
            joinColumns=@JoinColumn(name="personId"),
            inverseJoinColumns=@JoinColumn(name="friendId")
    )
    private List<User> friends;
}

用户-朋友映射表只有两列，这两列都是tbl_users表的uid列的外键。 两列分别是personId （应该映射到当前用户）和friendId （指定当前用户的朋友的id）。

问题是，即使我已经预先填充了Friends表，使得系统中的所有用户都是与所有其他用户的朋友，“ friends”字段仍然保持为空。 我什至尝试将关系切换为@OneToMany ，但仍然显示为空（尽管Hibernate调试输出显示SELECT * FROM tbl_friends WHERE personId = ? AND friendId = ?查询，但SELECT * FROM tbl_friends WHERE personId = ? AND friendId = ?没有其他）。

关于如何填充此列表的任何想法？ 谢谢！

Answer 1

@ManyToMany对自己来说相当令人困惑，因为您通常的建模方式不同于“休眠”方式。 您的问题是您缺少另一个收藏夹。

可以这样考虑-如果要“多对多”映射“作者” /“书”，则需要“书”上的“作者”集合和“作者”上的“书”集合。 在这种情况下，您的“用户”实体代表关系的两端。 因此您需要“我的朋友”和“朋友的”收藏集：

@ManyToMany
@JoinTable(name="tbl_friends",
 joinColumns=@JoinColumn(name="personId"),
 inverseJoinColumns=@JoinColumn(name="friendId")
)
private List<User> friends;

@ManyToMany
@JoinTable(name="tbl_friends",
 joinColumns=@JoinColumn(name="friendId"),
 inverseJoinColumns=@JoinColumn(name="personId")
)
private List<User> friendOf;

您仍然可以使用相同的关联表，但是请注意，join / inverseJon列在集合上交换。

“ friends”和“ friendOf”集合可能匹配，也可能不匹配（取决于您的“友谊”是否始终是相互的），您当然不必在API中以这种方式公开它们，但这就是映射的方式它在休眠状态。

Answer 2

实际上，它非常简单，可以通过以下方式实现：

public class Human {
    int id;
    short age;
    String name;
    List<Human> relatives;



    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public short getAge() {
        return age;
    }
    public void setAge(short age) {
        this.age = age;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public List<Human> getRelatives() {
        return relatives;
    }
    public void setRelatives(List<Human> relatives) {
        this.relatives = relatives;
    }

    public void addRelative(Human relative){
        if(relatives == null)
            relatives = new ArrayList<Human>();
        relatives.add(relative);
    }




}

HBM相同：

<hibernate-mapping>
    <class name="org.know.july31.hb.Human" table="Human">
        <id name="id" type="java.lang.Integer">
            <column name="H_ID" />
            <generator class="increment" />
        </id>
        <property name="age" type="short">
            <column name="age" />
        </property>
        <property name="name" type="string">
            <column name="NAME" length="200"/>
        </property>
        <list name="relatives" table="relatives" cascade="all">
         <key column="H_ID"/>
         <index column="U_ID"/>
         <many-to-many class="org.know.july31.hb.Human" column="relation"/>
      </list>
    </class>
</hibernate-mapping>

和测试用例

import org.junit.Test;
import org.know.common.HBUtil;
import org.know.july31.hb.Human;

public class SimpleTest {

    @Test
    public void test() {
        Human h1 = new Human();
        short s = 23;
        h1.setAge(s);
        h1.setName("Ratnesh Kumar singh");
        Human h2 = new Human();
        h2.setAge(s);
        h2.setName("Praveen Kumar singh");
        h1.addRelative(h2);
        Human h3 = new Human();
        h3.setAge(s);
        h3.setName("Sumit Kumar singh");
        h2.addRelative(h3);
        Human dk = new Human();
        dk.setAge(s);
        dk.setName("D Kumar singh");
        h3.addRelative(dk);
        HBUtil.getSessionFactory().getCurrentSession().beginTransaction();
        HBUtil.getSessionFactory().getCurrentSession().save(h1);
        HBUtil.getSessionFactory().getCurrentSession().getTransaction().commit();
        HBUtil.getSessionFactory().getCurrentSession().beginTransaction();
        h1 = (Human)HBUtil.getSessionFactory().getCurrentSession().load(Human.class, 1);
        System.out.println(h1.getRelatives().get(0).getName());

        HBUtil.shutdown();
    }

}

Answer 3

@JoinTable批注似乎使所接受的答案过于复杂。 稍微简单一些的实现只需要一个mappedBy 。 使用mappedBy表示拥有的Entity或属性，它可能应该是referencesTo因为这将被视为“朋友”。 ManyToMany关系可以创建非常复杂的图。 使用mappedBy可以使代码如下：

@Entity
public class Recursion {
    @Id @GeneratedValue
    private Integer id;
    // what entities does this entity reference?
    @ManyToMany
    private Set<Recursion> referencesTo;
    // what entities is this entity referenced from?
    @ManyToMany(mappedBy="referencesTo")
    private Set<Recursion> referencesFrom;
    public Recursion init() {
        referencesTo = new HashSet<>();
        return this;
    }
    // getters, setters
}

要使用它，您需要考虑拥有属性是referencesTo 。 您只需要在该属性中放置关系即可对其进行引用。 当您读回Entity ，假设您执行了fetch join ，则JPA将为结果创建集合。 删除实体时，JPA会删除对该实体的所有引用。

tx.begin();
Recursion r0 = new Recursion().init();
Recursion r1 = new Recursion().init();
Recursion r2 = new Recursion().init();
r0.getReferencesTo().add(r1);
r1.getReferencesTo().add(r2);
em.persist(r0);
em.persist(r1);
em.persist(r2);

tx.commit();
// required so that existing entities with null referencesFrom will be removed from cache.
em.clear();
for ( int i=1; i <= 3; ++i ) {
    Recursion r = em.createQuery("select distinct r from Recursion r left join fetch r.referencesTo left join fetch r.referencesFrom where id = :id", Recursion.class).setParameter("id",  i).getSingleResult();
    System.out.println(r + " To=" + Arrays.toString(r.getReferencesTo().toArray()) + " From=" + Arrays.toString(r.getReferencesFrom().toArray()) );
}
tx.begin();
em.createQuery("delete from Recursion where id = 2").executeUpdate();
tx.commit();
// required so that existing entities with referencesTo will be removed from cache.
em.clear();
Recursion r = em.createQuery("select distinct r from Recursion r left join fetch r.referencesTo left join fetch r.referencesFrom where id = :id", Recursion.class).setParameter("id",  1).getSingleResult();
System.out.println(r + " To=" + Arrays.toString(r.getReferencesTo().toArray()) + " From=" + Arrays.toString(r.getReferencesFrom().toArray()) );

它提供以下日志输出（始终检查生成的SQL语句）：

Hibernate: create table Recursion (id integer not null, primary key (id))
Hibernate: create table Recursion_Recursion (referencesFrom_id integer not null, referencesTo_id integer not null, primary key (referencesFrom_id, referencesTo_id))
Hibernate: create sequence hibernate_sequence start with 1 increment by 1
Hibernate: alter table Recursion_Recursion add constraint FKsi0wfuwfs0bl19jjpofw4n8pt foreign key (referencesTo_id) references Recursion
Hibernate: alter table Recursion_Recursion add constraint FKarrkuyh2v1j5qnlui2vbpl7tk foreign key (referencesFrom_id) references Recursion
Hibernate: call next value for hibernate_sequence
Hibernate: call next value for hibernate_sequence
Hibernate: call next value for hibernate_sequence
Hibernate: insert into Recursion (id) values (?)
Hibernate: insert into Recursion (id) values (?)
Hibernate: insert into Recursion (id) values (?)
Hibernate: insert into Recursion_Recursion (referencesFrom_id, referencesTo_id) values (?, ?)
Hibernate: insert into Recursion_Recursion (referencesFrom_id, referencesTo_id) values (?, ?)
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@7bdf6bb7 To=[model.Recursion@1bc53649] From=[]
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@1bc53649 To=[model.Recursion@42deb43a] From=[model.Recursion@7bdf6bb7]
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@42deb43a To=[] From=[model.Recursion@1bc53649]
Hibernate: delete from Recursion_Recursion where (referencesTo_id) in (select id from Recursion where id=2)
Hibernate: delete from Recursion_Recursion where (referencesFrom_id) in (select id from Recursion where id=2)
Hibernate: delete from Recursion where id=2
Hibernate: select distinct recursion0_.id as id1_2_0_, recursion2_.id as id1_2_1_, recursion4_.id as id1_2_2_, references1_.referencesFrom_id as referenc1_3_0__, references1_.referencesTo_id as referenc2_3_0__, references3_.referencesTo_id as referenc2_3_1__, references3_.referencesFrom_id as referenc1_3_1__ from Recursion recursion0_ left outer join Recursion_Recursion references1_ on recursion0_.id=references1_.referencesFrom_id left outer join Recursion recursion2_ on references1_.referencesTo_id=recursion2_.id left outer join Recursion_Recursion references3_ on recursion0_.id=references3_.referencesTo_id left outer join Recursion recursion4_ on references3_.referencesFrom_id=recursion4_.id where id=?
model.Recursion@6b739528 To=[] From=[]