另一个Hibernate问题......：P
使用Hibernate的Annotations框架，我有一个User实体。每个User都可以拥有一组朋友：其他User的集合。但是，我无法弄清楚如何在由User列表组成的User类中创建多对多关联（使用用户朋友中间表）。 / p>
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;
}
用户友好映射表只有两列，这两列都是uid表的tbl_users列的外键。这两列是personId（应该映射到当前用户），friendId（指定当前用户的朋友的ID）。
问题是，“friends”字段一直显示为null，即使我已预先填充了friends表，使得系统中的所有用户都是所有其他用户的朋友。我甚至尝试将关系切换到@OneToMany，它仍然是null（尽管Hibernate调试输出显示SELECT * FROM tbl_friends WHERE personId = ? AND friendId = ?查询，但没有其他内容。
有关如何填充此列表的任何想法？谢谢！
答案 0 :(得分：60)
以这种方式思考 - 如果您将“作者”/“书籍”映射为多对多，则需要书籍上的“作者”集合和作者的“书籍”集合。在这种情况下，您的“用户”实体代表关系的两端;所以你需要“我的朋友”和“朋友”系列：
@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，但请注意，在集合上交换join / inverseJon列。
“friends”和“friendOf”系列可能匹配也可能不匹配（取决于您的“友谊”是否始终相互），当然，您不必在API中以这种方式公开它们，但这是在Hibernate中映射它的方法。
答案 1 :(得分：0)
实际上它很简单，可以通过以下说明你有以下实体来实现
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();
}
}
答案 2 :(得分：0)
接受的答案似乎过于复杂@JoinTable注释。稍微简单的实现只需要mappedBy。使用mappedBy表示拥有Entity或属性，它应该是referencesTo，因为它将被视为“朋友”。 ManyToMany关系可以创建非常复杂的图表。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，以便引用它们。当您阅读Entity时，假设您执行fetch join，JPA将为结果创建集合。删除实体时，JPA将删除对它的所有引用。
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=[]