Question

我是JPA的新手并做了一些小样本来了解它。但我在下面遇到一个问题，请帮帮我，请解释原因：

我有类Customer.java，它在db：

中映射到表customer

@Entity
public class Customer implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @Column(name = "id_customer")
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    private Long id;

    // accountNumber field maps with accountNumber column in Account table
    @Column(name = "loginId", unique = true)
    private String loginId;

    @Column(name = "password")
    private String password;

    @Column(name = "firstName")
    private String firstName;

    @Column(name = "lastName")
    private String lastName;

    @Column(name = "address")
    private String address;

    @Column(name = "email")
    private String email;

    @Column(name = "phone")
    private String phone;

    @OneToMany(mappedBy="customer")
    private List<Account> accountList;

    @OneToMany(mappedBy="customer")
    private List<Card> cardList;
// getters and setters goes here
}

上面的类有两个列表，accountList和cardList，它们的泛型类（Card和Account）使用单表继承扩展BaseInfo。
这是我的BaseInfo.java：

@Entity
@Table(name = "account")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "discriminator", discriminatorType = DiscriminatorType.STRING)
public class BaseInfo implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @Column(name = "number")
    private String number;

    @Column(name = "availableNumber")
    private Long availableNumber;
//getter and setter here
}

Class Card.java：

@Entity
@Table(name = "account")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorValue(value = "C")
public class Card extends BaseInfo implements Serializable {

    private static final long serialVersionUID = 1L;

    @Column(name = "cardType")
    private String cardType;

    @ManyToOne
    @JoinColumn(name = "id_customer")
    private Customer customer;

//getter and setter
}

并且类Account.java：

@Entity
@Table(name = "account")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorValue(value = "A")
public class Account extends BaseInfo implements Serializable {
    private static final long serialVersionUID = 1L;

    @Column(name = "accountName")
    private String accountName;

    @Column(name = "accountType")
    private String accountType;

    @Temporal(TemporalType.TIMESTAMP)
    @Column(name = "dt_created")
    private Date createdDate;

    @Temporal(TemporalType.TIMESTAMP)
    @Column(name = "dt_lst_updt")
    private Date lastUpdatedDate;

    @ManyToOne
    @JoinColumn(name = "id_customer")
    private Customer customer;

//getter, setter
}

然后，我执行查询，使用loginid和密码从数据库查询客户，如下所示：

entityTransaction.begin();

TypedQuery<Customer> query = entityManager.createQuery(
                    "SELECT c FROM " + Customer.class.getName()
                            + " c Where c.loginId= :loginId", Customer.class);
query.setParameter("loginId", loginId);

res = query.getSingleResult();

entityTransaction.commit();

代码运行没有错误，但结果对我来说很奇怪：当我调试（或将结果打印到jsp）时，accountList或cardList包含该客户的所有帐户，就像他们不关心“鉴别者”专栏。

我有两个问题：

如何归档listCard仅包含Card（歧视= c）且listAccount仅包含Account（discriminator = a）的目标？
是否有另一种方法可以查询listCard或listAccount而无需先查询客户（就像我使用的那样）??

提前感谢！：d

Answer 1

我不确定它是JPA限制还是特定于Hibernate的限制，但您可能不会使用相同的列来映射两个不同的关联。

您应该使用类似car_customer_id的内容来映射客户和卡之间的关联，并使用account_customer_id来映射客户和帐户之间的关联。

了解JPA中的单表继承

1 个答案: