我有3张表与多对多的关系
我可以为hibernate建模类,以便我使用 IdClass 而不是 EmbeddedId 解决方案(这样我就可以避免使用级联不必要的级别来调用 - getId()。getA ())
PS尝试混合这些解决方案,但它不起作用:答案 0 :(得分:6)
当谈到建立实体之间的关系时,我旁边最近的墙和我的头经常加入......(斯蒂芬)
以下是实体A和B之间多对多关系的工作示例:
<强> A.java 强>
import java.util.ArrayList;
import java.util.List;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
@Entity
public class A {
@Id
@GeneratedValue(strategy=GenerationType.SEQUENCE)
@Column(name="id_a")
private Integer id;
private String name;
@OneToMany(mappedBy="a",cascade=CascadeType.ALL, fetch=FetchType.LAZY)
private List<AB> abAssociations = new ArrayList<>();
// Getters and setters...
}
<强> B.java 强>
import java.util.ArrayList;
import java.util.List;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
@Entity
public class B {
@Id
@GeneratedValue(strategy=GenerationType.SEQUENCE)
@Column(name = "id_b")
private Integer id;
private String name;
@OneToMany(mappedBy = "b", cascade=CascadeType.ALL, fetch=FetchType.LAZY)
private List<AB> abAssociations = new ArrayList<>();
// Getters and setters...
}
<强> AB.java 强>
import java.util.Date;
import javax.persistence.CascadeType;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.Id;
import javax.persistence.IdClass;
import javax.persistence.ManyToOne;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;
@Entity
@IdClass(ABid.class)
public class AB {
@Id
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.LAZY)
private A a;
@Id
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.LAZY)
private B b;
@Temporal(TemporalType.TIMESTAMP)
private Date date;
// Getters and setters...
}
<强> ABid.java 强>
import java.io.Serializable;
// The IdClass MUST implement Serializable and override #hashCode and #equals
public class ABid implements Serializable {
private static final long serialVersionUID = -2834827403836993112L;
private A a;
private B b;
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((a == null) ? 0 : a.hashCode());
result = prime * result + ((b == null) ? 0 : b.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
ABid other = (ABid) obj;
if (a == null) {
if (other.a != null)
return false;
} else if (!a.equals(other.a))
return false;
if (b == null) {
if (other.b != null)
return false;
} else if (!b.equals(other.b))
return false;
return true;
}
}
<强>的pom.xml 强>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<version>4.3.7.Final</version>
</dependency>
import java.util.Date;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
public class Main {
public static void main(String[] args) {
// * Init entity manager
EntityManagerFactory emf = Persistence.createEntityManagerFactory("playground");
EntityManager em = emf.createEntityManager();
em.getTransaction().begin();
// * Create two entities and persist them
// We must persist the entities first alone before we build and flush their relation
A a = new A();
a.setName("foo");
em.persist(a);
B b = new B();
b.setName("bar");
em.persist(b);
// * Build relationships between the two previous entities
AB ab = new AB();
ab.setA(a);
ab.setB(b);
ab.setDate(new Date());
a.getAbAssociations().add(ab);
b.getAbAssociations().add(ab);
// * Flush our changements in the database
em.getTransaction().commit();
em.close();
emf.close();
}
}
这是Hibernate在Postgresql数据库上创建的表的sql代码。
CREATE TABLE a
(
id_a integer NOT NULL,
name character varying(255),
CONSTRAINT a_pkey PRIMARY KEY (id_a)
)
CREATE TABLE b
(
id_b integer NOT NULL,
name character varying(255),
CONSTRAINT b_pkey PRIMARY KEY (id_b)
)
CREATE TABLE ab
(
date timestamp without time zone,
b_id_b integer NOT NULL,
a_id_a integer NOT NULL,
CONSTRAINT ab_pkey PRIMARY KEY (a_id_a, b_id_b),
CONSTRAINT fk_3exna7nsxvj1kv9i9pntmwlf1 FOREIGN KEY (a_id_a)
REFERENCES a (id_a) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT fk_n3jrq53nr1elew4rytocopkbu FOREIGN KEY (b_id_b)
REFERENCES b (id_b) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
答案 1 :(得分:1)
回应@valik的评论。
尝试一下:
@Entity
public class AB {
@Id
@GeneratedValue(strategy=GenerationType.SEQUENCE)
@Column(name = "id_ab")
private Integer id;
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.LAZY)
private A a;
@ManyToOne(cascade=CascadeType.ALL, fetch=FetchType.LAZY)
private B b;
@Temporal(TemporalType.TIMESTAMP)
private Date date;
// Getters and setters...
}