对非主键列的oneToone关系
-
21-12-2019 - |
题
当我查询与另一个有oneoone关系的实体时,我遇到了问题。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");
...
.
de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=?
.
正如您所看到的,加入是错误的,因为正在与人们比较用户.ID( user1_.email= person2_id ),因此它返回一个用户没有相应的人。
关于如何修复它的任何想法?
非常感谢!!
解决方案
我认为你应该重新考虑你的DataModel。用户和人与人之间的关系看起来更像是继承之一。
在您的映射时,他们认为在这里看到一些进一步的混乱:
其他提示
严格来说, JPA规范不允许对非主键列的引用。它可能在一些JPA实现中工作,但它不合适。
但是,我认为您可以通过使关系双向的关系,并由非主键拥有:@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;
}
.
我没有实际尝试过,所以警告黑客。
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