[英]Understanding Single Table Inheritance in JPA
我是JPA的新手,并且做了一个小样本来了解它。 但是我在下面遇到一个问题,请帮帮我,并解释原因:
我有类Customer.java,它映射到db中的表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,它们的通用类(Card和Account)使用Single table Inheritance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
}
Card.java类:
@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
}
和类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,我执行一个查询,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,但是结果令我有些奇怪:当我调试(或将结果打印到jsp)时,accountList或cardList包含该客户的所有Account,就像他们不关心'discriminator' '列。
我有两个问题:
如何归档listCard仅包含Card(区分= c)和listAccount仅包含Account(discriminator = a)的目标?
有没有另外一种查询listCard或listAccount的方法,而无需先查询客户(例如我使用的)?
预先感谢! :D
我不确定这是JPA限制还是特定于Hibernate的限制,但您可能无法使用同一列来映射两个不同的关联。
您应该使用诸如car_customer_id
类的car_customer_id
来映射客户与卡之间的关联,并使用account_customer_id
来映射客户与帐户之间的关联。
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