java - 非主键列的一对一关系

Question

当我查询与另一个实体具有 OneToOne 关系的实体时遇到问题。这是场景:

数据库表:

create table people (
    id decimal(10,0) NOT NULL,
    email varchar(512) NOT NULL
);

create table users (
    email varchar(512) NOT NULL
);

测试数据:

insert into users (email) values ('jhon@domain.com');
insert into users (email) values ('mary@domain.com');

insert into people (id, email) values (1, 'jhon@domain.com');
insert into people (id, email) values (2, 'mary@domain.com');

实体:

@Entity(name = "people")
public class Person implements Serializable {

    @Column
    @Id
    private long id;

    @Column
    private String email;

    public long getId() {
        return id;
    }

    public void setId(long id) {
        this.id = id;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

}

@Entity(name = "tbl_users")
public class User implements Serializable {

    @Id
    private String email;

    @OneToOne(cascade=CascadeType.ALL, fetch=FetchType.EAGER)
    @JoinColumn(name = "email", referencedColumnName = "email")
    private Person person;

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }
}

调用:

...
User user = entityManager.find(User.class, "jhon@domain.com");
...

取消调用后，hibernate 的日志显示:

select user1_.email as email2_0_, person2_.id as id1_1_, person2_.email as email1_1_
from users user1_ left outer join people person2_ on user1_.email=person2_.id
where user1_.email=?

如您所见，连接是错误的，因为比较 users.email 和 people.id (user1_.email=person2_.id)，因此它返回一个 User 没有相应的人。

关于如何修复它有什么想法吗？

非常感谢!!

Answer 1

严格来说，the JPA specification does not allow references to non-primary key columns .它可能适用于某些 JPA 实现，但并不合适。

但是，我认为您可以通过使关系成为双向的，并由具有非主键的一方拥有来做到这一点:

@Entity
public class Person {

    @Id
    private long id;

    @OneToOne
    @JoinColumn(name = "email")
    private User user;

    public String getEmail() {
        return user.getEmail();
    }

    public void setEmail(String email) {
        // left as an exercise for the reader
    }

}

@Entity
public class User {

    @Id
    private String email;

    @OneToOne(mappedBy = "user")
    private Person person;

}

不过，我还没有实际尝试过，所以警告黑客。