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[英]Hibernate count NamedQuery for many-to-many association with the same entity
[英]Hibernate recursive many-to-many association with the same entity
另一個休眠問題...:P
使用Hibernate的Annotations框架,我有一個User
實體。 每個User
可以有一個朋友集合:其他User
的集合。 但是,我還無法弄清楚如何在User
類中創建一個由User
列表組成的多對多關聯(使用用戶友好中間表)。
這是User類及其注釋:
@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;
}
用戶-朋友映射表只有兩列,這兩列都是tbl_users
表的uid
列的外鍵。 兩列分別是personId
(應該映射到當前用戶)和friendId
(指定當前用戶的朋友的id)。
問題是,即使我已經預先填充了Friends表,使得系統中的所有用戶都是與所有其他用戶的朋友,“ friends”字段仍然保持為空。 我什至嘗試將關系切換為@OneToMany
,但仍然顯示為空(盡管Hibernate調試輸出顯示SELECT * FROM tbl_friends WHERE personId = ? AND friendId = ?
查詢,但SELECT * FROM tbl_friends WHERE personId = ? AND friendId = ?
沒有其他)。
關於如何填充此列表的任何想法? 謝謝!
@ManyToMany對自己來說相當令人困惑,因為您通常的建模方式不同於“休眠”方式。 您的問題是您缺少另一個收藏夾。
可以這樣考慮-如果要“多對多”映射“作者” /“書”,則需要“書”上的“作者”集合和“作者”上的“書”集合。 在這種情況下,您的“用戶”實體代表關系的兩端。 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關聯表,但是請注意,join / inverseJon列在集合上交換。
“ friends”和“ friendOf”集合可能匹配,也可能不匹配(取決於您的“友誼”是否始終是相互的),您當然不必在API中以這種方式公開它們,但這就是映射的方式它在休眠狀態。
實際上,它非常簡單,可以通過以下方式實現:
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>
和測試用例
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();
}
}
@JoinTable
批注似乎使所接受的答案過於復雜。 稍微簡單一些的實現只需要一個mappedBy
。 使用mappedBy
表示擁有的Entity
或屬性,它可能應該是referencesTo
因為這將被視為“朋友”。 ManyToMany
關系可以創建非常復雜的圖。 使用mappedBy
可以使代碼如下:
@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,您需要考慮擁有屬性是referencesTo
。 您只需要在該屬性中放置關系即可對其進行引用。 當您讀回Entity
,假設您執行了fetch join
,則JPA將為結果創建集合。 刪除實體時,JPA會刪除對該實體的所有引用。
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()) );
它提供以下日志輸出(始終檢查生成的SQL語句):
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=[]
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