java - 注释 m :n relationship with helper class (2 foreign keys + additional attributes) with JPA 2. 0

Question

我是 JPA 2.0 的新手，在注释 n:m 关系时遇到了麻烦，该关系使用外键类和描述关系的附加属性:

客户可以订阅多本杂志，每期Subscription正好为一个 Customer 创建和一个 Magazine ，另外节省订阅的持续时间。

这是我的带注释的类，我正在使用字段访问。我省略了一些样板代码，如构造函数、setter、getter(未注释)和方法 toString , equals和 hashCode .

@Entity
public class Customer {
    @Id
    @GeneratedValue
    private Long id;
    private String name;
    @ManyToMany
    private Set<Subscription> subscriptions;

    // ..
 }

@Entity
public class Magazine {
    @Id
    @GeneratedValue
    private Long id;
    private String name;

    @ManyToMany
    private Set<Subscription> subscriptions;

    // ..
}

@Entity
public class Subscription {
    private Date start;
    private Date end;

    @EmbeddedId
    private SubscriptionId id;

    // ..
}

@Embeddable
public class SubscriptionId implements Serializable {
    @ManyToOne
    private Customer customer;
    @ManyToOne
    private Magazine magazine;

    // ..
}

我通过创建和保留一些对象来测试我的注释，如下所示:

public static void main(String[] args) {
    EntityManagerFactory emf = Persistence
            .createEntityManagerFactory("kiosk");
    EntityManager em = emf.createEntityManager();

    persist(em);

    em.close();
    emf.close();
}

private static void persist(EntityManager em) {
    em.getTransaction().begin();

    Magazine mag1 = new Magazine("mag1");
    Magazine mag2 = new Magazine("mag2");

    Customer cus1 = new Customer("cus1");
    Customer cus2 = new Customer("cus2");
    Customer cus3 = new Customer("cus3");

    Subscription sub1 = new Subscription(cus1, mag1);
    Subscription sub2 = new Subscription(cus2, mag1);
    Subscription sub3 = new Subscription(cus2, mag2);
    Subscription sub4 = new Subscription(cus3, mag2);

    em.persist(mag1);
    em.persist(mag2);

    em.persist(cus1);
    em.persist(cus2);
    em.persist(cus3);

    em.persist(sub1);
    em.persist(sub2);
    em.persist(sub3);
    em.persist(sub4);

    em.getTransaction().commit();
}

提供程序创建以下 MySQL 数据库:

mysql> show tables;
+-----------------------+
| Tables_in_kiosk       |
+-----------------------+
| customer              |
| customer_subscription |
| magazine              |
| magazine_subscription |
| subscription          |
+-----------------------+

只有三个表 customer , magazine和 subscription有内容:

mysql> select * from customer;
+-------------+------+
| customer_id | name |
+-------------+------+
|           1 | cus1 |
|           2 | cus2 |
|           3 | cus3 |
+-------------+------+

mysql> select * from magazine;
+-------------+------+
| magazine_id | name |
+-------------+------+
|           1 | mag1 |
|           2 | mag2 |
+-------------+------+

mysql> select * from subscription;
+------+-------+-------------+-------------+
| end  | start | magazine_id | customer_id |
+------+-------+-------------+-------------+
| NULL | NULL  |           1 |           1 |
| NULL | NULL  |           1 |           2 |
| NULL | NULL  |           2 |           2 |
| NULL | NULL  |           2 |           3 |
+------+-------+-------------+-------------+

如果我知道他们的 key ，我就可以阅读我的订阅。不过，我还没有尝试为客户或杂志阅读整套内容。

private static void find(EntityManager em) {
    Magazine mag1 = em.find(Magazine.class, 1L);
    Magazine mag2 = em.find(Magazine.class, 2L);

    Customer cus1 = em.find(Customer.class, 1L);
    Customer cus2 = em.find(Customer.class, 2L);
    Customer cus3 = em.find(Customer.class, 3L);

    Subscription sub1 = em.find(Subscription.class, new SubscriptionId(cus1, mag1));
    Subscription sub2 = em.find(Subscription.class, new SubscriptionId(cus2, mag1));
    Subscription sub3 = em.find(Subscription.class, new SubscriptionId(cus2, mag2));
    Subscription sub4 = em.find(Subscription.class, new SubscriptionId(cus3, mag2));

    System.out.println(mag1);
    System.out.println(mag2);

    System.out.println(cus1);
    System.out.println(cus2);
    System.out.println(cus3);

    System.out.println(sub1);
    System.out.println(sub2);
    System.out.println(sub3);
    System.out.println(sub4);
}

打印:

Magazine [id=1, name=mag1, subscriptions=null]
Magazine [id=2, name=mag2, subscriptions=null]
Customer [id=1, name=cus1, subscriptions=null]
Customer [id=2, name=cus2, subscriptions=null]
Customer [id=3, name=cus3, subscriptions=null]
Subscription [start=null, end=null, id=SubscriptionId [customer=1, magazine=1]]
Subscription [start=null, end=null, id=SubscriptionId [customer=2, magazine=1]]
Subscription [start=null, end=null, id=SubscriptionId [customer=2, magazine=2]]
Subscription [start=null, end=null, id=SubscriptionId [customer=3, magazine=2]]

两张表 customer_subscription和 magazine_subscription保持空。但在我看来，它们甚至没有必要——其他 3 个表看起来和我想要的完全一样。所以我的问题是:

如何使用 JPA 2.0 正确建模此示例中使用的 m:n 关系，同时不创建多余的表，同时保留编写和读取杂志或客户的所有订阅的能力？

如果有人对代码感兴趣，我已将其上传到此处: http://goo.gl/qSc2e ;您需要一个名为“kiosk”的 MySQL 5 数据库，该数据库在本地主机上的端口 3306 上运行，root 密码为空。

Answer 1

您的模型实际上没有任何多对多关系。
Customer至 Subscription是一对多。Magazine至 Subscription是一对多。

试试这个作为你的实体模型:

@Entity
public class Customer {

    @Id
    @GeneratedValue
    private Long id;

    private String name;

    @OneToMany
    private Set<Subscription> subscriptions;

    // ..
 }

@Entity
public class Magazine {
    @Id
    @GeneratedValue
    private Long id;

    private String name;

    @OneToMany
    private Set<Subscription> subscriptions;

    // ..
}

@Entity
public class Subscription {

    private Date start;

    private Date end;

    @EmbeddedId
    private SubscriptionId id;

    // ..
}

@Embeddable
public class SubscriptionId implements Serializable {

    @ManyToOne
    private Customer customer;

    @ManyToOne
    private Magazine magazine;

    // ..
}

并且 Hibernate 不会生成冗余链接表。