【发布时间】:2014-10-03 21:04:58
【问题描述】:
根据Hibernate documentation,@MapsId注解的解释如下:
在嵌入的 id 对象中,关联表示为 关联实体的标识符。但是您可以将其值链接到 通过 @MapsId 注释在实体中进行常规关联。这 @MapsId 值对应嵌入id的属性名 包含关联实体标识符的对象。在数据库中, 这意味着 Customer.user 和 CustomerId.userId 属性 共享相同的基础列(在本例中为 user_fk)。
@Entity
class Customer {
@EmbeddedId CustomerId id;
boolean preferredCustomer;
@MapsId("userId")
@JoinColumns({
@JoinColumn(name="userfirstname_fk", referencedColumnName="firstName"),
@JoinColumn(name="userlastname_fk", referencedColumnName="lastName")
})
@OneToOne User user;
}
@Embeddable
class CustomerId implements Serializable {
UserId userId;
String customerNumber;
//implements equals and hashCode
}
@Entity
class User {
@EmbeddedId UserId id;
Integer age;
}
@Embeddable
class UserId implements Serializable {
String firstName;
String lastName;
//implements equals and hashCode
}
它还说:
虽然 JPA 不支持,但 Hibernate 允许您放置关联 直接在嵌入的 id 组件中(而不必使用 @MapsId 注解)。
@Entity
class Customer {
@EmbeddedId CustomerId id;
boolean preferredCustomer;
}
@Embeddable
class CustomerId implements Serializable {
@OneToOne
@JoinColumns({
@JoinColumn(name="userfirstname_fk", referencedColumnName="firstName"),
@JoinColumn(name="userlastname_fk", referencedColumnName="lastName")
})
User user;
String customerNumber;
//implements equals and hashCode
}
@Entity
class User {
@EmbeddedId UserId id;
Integer age;
}
@Embeddable
class UserId implements Serializable {
String firstName;
String lastName;
//implements equals and hashCode
}
我尝试使用 Hibernate 本身 (hbm2ddl.auto=create) 生成表,以了解如何使用 @MapsId 注释。以下是我的观察:
如果我对Customer 和User 的实体声明是这样的:
@Entity
@Table(name="TBL_CUSTOMER")
public class Customer {
@EmbeddedId CustomerId id;
boolean preferredCustomer;
@MapsId("userId")
@JoinColumns({
@JoinColumn(name="userfirstname_fk", referencedColumnName="firstName"),
@JoinColumn(name="userlastname_fk", referencedColumnName="lastName")
})
@OneToOne User user;
}
@Entity
@Table(name="TBL_USER")
class User {
@EmbeddedId UserId id;
Integer age;
}
然后Hibernate生成的DDL语句说:
Hibernate: create table TBL_CUSTOMER (customerNumber varchar2(255 char) not null, preferredCustomer number(1,0) not null, userfirstname_fk varchar2(255 char) not null, userlastname_fk varchar2(255 char) not null, primary key (customerNumber, userfirstname_fk, userlastname_fk))
Hibernate: create table TBL_USER (firstName varchar2(255 char) not null, lastName varchar2(255 char) not null, age number(10,0), primary key (firstName, lastName))
Hibernate: alter table TBL_CUSTOMER add constraint UK_chvh5mukc81xk9t6fis3skab unique (userfirstname_fk, userlastname_fk)
Hibernate: alter table TBL_CUSTOMER add constraint FK_chvh5mukc81xk9t6fis3skab foreign key (userfirstname_fk, userlastname_fk) references TBL_USER
现在,如果我将我的 Customer 实体更改为:
@Entity
@Table(name="TBL_CUSTOMER")
public class Customer {
@EmbeddedId CustomerId id;
boolean preferredCustomer;
@OneToOne User user;
}
那么 DDL 语句是:
Hibernate: create table TBL_CUSTOMER (customerNumber varchar2(255 char) not null, firstName varchar2(255 char), lastName varchar2(255 char), preferredCustomer number(1,0) not null, user_firstName varchar2(255 char), user_lastName varchar2(255 char), primary key (customerNumber, firstName, lastName))
Hibernate: create table TBL_USER (firstName varchar2(255 char) not null, lastName varchar2(255 char) not null, age number(10,0), primary key (firstName, lastName))
Hibernate: alter table TBL_CUSTOMER add constraint FK_et3bgekef237d4kov7b9oqt85 foreign key (user_firstName, user_lastName) references TBL_USER
在这种情况下,如果我删除 @MapsId 和 @JoinColumn 注释,我会看到 TBL_CUSTOMER 的 2 个额外列(名字和姓氏)。在这种情况下,也没有额外的 alter 命令。
我是 Hibernate 的新手,所以我很难理解 Hibernate 文档中给出的解释、@MapsId 的用途、我们何时必须使用它以及它如何影响底层数据库架构。
我也浏览了这篇 SO 帖子 - can someone please explain me @MapsId in hibernate?,但我无法获得有关此注释的明确信息。
【问题讨论】:
-
我只使用带有
Map的映射。所以我的映射ID是它的关键。有时,它比List或Set更方便