在JPA、Hibernate中仅使用表名和列名连接表

在JPA、Hibernate中仅使用表名和列名连接表

我有两个实体UserService中的users和AddressService中的addresses，现在我想将address表连接到users表，而不使用users实体，只使用users表和user\u id作为外键

@Entity
@DynamicUpdate
@Table(name = "users", schema = "security", uniqueConstraints = { @UniqueConstraint(columnNames = { "email" }) })
@JsonPropertyOrder({ "user_id" })
public class User implements Serializable {


@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "user_id")
private Long userId;
private String email;
private String password;

//getters and setters
}

@Entity
@Table(name = "address", schema = "account")
@SecondaryTable(name = "users", schema = "account", pkJoinColumns = { @PrimaryKeyJoinColumn(name = "id") })
@JsonPropertyOrder({ "id" })
@DynamicUpdate
public class AddressModel implements Serializable {

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

    @NotNull
    @Valid
    @Embedded
    private Address address;

@ManyToOne(cascade = CascadeType.ALL)
@JoinTable(name = "users", schema = "account", joinColumns = @JoinColumn(name = "user_id", referencedColumnName = "user_id"))
public Long user_id;  // this is the primary key in users table 
                      // and want to make foreign key in addresses table

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这两个实体位于两个不同的RESTful web服务中。如何在不创建实体或不具有依赖关系的情况下连接这两列？

这会产生什么结果？您加入了表，但不想得到实体，那么结果应该是什么？为什么在一个长字段上有多个注释？所有JPA文档都会简单地告诉您关系是实体类型。如果该用户_id代表一个用户，则输入一个user类型的字段。面向对象再简单不过了