Hibernate(十一):映射继承关系的三种方案

时间:2023-03-10 06:08:29
Hibernate(十一):映射继承关系的三种方案
  • 背景:

  在一些项目中,会采用集成的关系来定义数据库实体类,比如:人(Person)与学生(Student),学生来源与人,所以人的基本属性学生也拥有;但学生有的一些属性,人就不具有。人与学生之间很显然就拥有了继承关系------学生继承于人,人是父类,学生是子类。

  那么,这种继承关系在hibernate是如何映射呢?

  对于面向对象的程序设计语言而言,继承和多态是两个最基本的概念。hibernate的集成映射可以理解为持久化类之间的继承关系。在上边的例子中,学生集成了人,可以认为学生是一个特殊的人,如果对人进行查询,学生的实例也将被得到。

  在hibernate中支持三种继承映射策略:

  1)使用subclass进行映射:将域模型中的每一个实体对象映射到一个独立的表中,也就是说不用在关系数据模型中考虑域模型中的继承关系和多态。

  2)使用jioned-subclass进行映射:对于继承关系中的子类使用同一个表,这就需要在数据库表中增加额外的区分子类类型的字段。

  3)使用union-subclass进行映射:域模型中的每个类映射到一个表,通过关系数据模型中的外键来描述表之间的继承关系。这也就相当于按照域模型的结构来建立数据库中的表,并通过外键来建立表之间的继承关系。

  • 使用subclass进行映射

新建工程Hibernate08,导入hibernate开发包及mysql驱动包。

在src下新建hibernate.cfg.xml配置文件:

 <?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
<property name="hibernate.connection.username">root</property>
<property name="hibernate.connection.password">123456</property>
<property name="hibernate.connection.driver_class">com.mysql.jdbc.Driver</property>
<property name="hibernate.connection.url">jdbc:mysql://localhost/hibernate_01</property> <!-- <property name="hibernate.dialect">org.hibernate.dialect.MySQLDialect</property>
<property name="hibernate.dialect">org.hibernate.dialect.MySQLInnoDBDialect</property> -->
<property name="hibernate.dialect">org.hibernate.dialect.MySQL5InnoDBDialect</property> <property name="hibernate.show_sql">true</property> <property name="hibernate.format_sql">true</property> <property name="hibernate.hbm2ddl.auto">update</property> <property name="hibernate.current_session_context_class">thread</property> <property name="hibernate.c3p0.max_size">500</property>
<property name="hibernate.c3p0.min_size">20</property>
<property name="hibernate.c3p0.max_statements">10</property>
<property name="hibernate.c3p0.timeout">2000</property>
<property name="hibernate.c3p0.idle_test_period">2000</property>
<property name="hibernate.c3p0.acquire_increment">10</property> <mapping resource="com/dx/hibernate06/extend/Person.hbm.xml" /> </session-factory>
</hibernate-configuration>

在src下新建com.dx.hibernate06.extend包,在该包下新建Person.java:

 package com.dx.hibernate06.extend;

 public class Person {
private Integer id;
private String name;
private Integer age; public Person() { } public Integer getId() {
return id;
} public void setId(Integer id) {
this.id = id;
} public String getName() {
return name;
} public void setName(String name) {
this.name = name;
} public Integer getAge() {
return age;
} public void setAge(Integer age) {
this.age = age;
}
}

新建Student.java(继承自Person):

 package com.dx.hibernate06.extend;

 public class Student extends Person {
private String className;
private String schoolName; public Student() { } public String getClassName() {
return className;
} public void setClassName(String className) {
this.className = className;
} public String getSchoolName() {
return schoolName;
} public void setSchoolName(String schoolName) {
this.schoolName = schoolName;
} }

添加Person.hbm.xml配置文件:

 <?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated 2017-6-8 15:16:00 by Hibernate Tools 3.5.0.Final -->
<hibernate-mapping package="com.dx.hibernate06.extend">
<class name="Person" table="PERSONS" discriminator-value="person">
<id name="id" type="java.lang.Integer">
<column name="ID" />
<generator class="native" />
</id> <!-- 配置辨别者列 -->
<discriminator column="TYPE" type="string"></discriminator> <property name="name" type="java.lang.String">
<column name="NAME" />
</property>
<property name="age" type="java.lang.Integer">
<column name="AGE" />
</property> <subclass name="Student" discriminator-value="student">
<property name="className" column="CLASS_NAME" type="string"></property>
<property name="schoolName" column="SCHOOL_NAME" type="string"></property>
</subclass>
</class>
</hibernate-mapping>

注意:

1)在这里并没有定义student.hbm.xml配置文件,而是只需要在Person.hbm.xml配置文件中设置subclass配置项就可以;

2)除了设置subclass节点,还需要添加discriminator(辨别列)节点配置,且需要在table和subclass节点中设置discriminator-value(辨别列值)。

添加测试类TestMain.java:

 package com.dx.hibernate06.extend;

 import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
import org.hibernate.boot.Metadata;
import org.hibernate.boot.MetadataSources;
import org.hibernate.boot.model.naming.ImplicitNamingStrategyComponentPathImpl;
import org.hibernate.boot.registry.StandardServiceRegistry;
import org.hibernate.boot.registry.StandardServiceRegistryBuilder;
import org.junit.After;
import org.junit.Before;
import org.junit.Test; public class TestMain {
private SessionFactory sessionFactory = null;
private Session session = null;
private Transaction transaction = null; @Before
public void init() {
StandardServiceRegistry standardRegistry = new StandardServiceRegistryBuilder().configure().build();
Metadata metadata = new MetadataSources(standardRegistry).getMetadataBuilder().applyImplicitNamingStrategy(ImplicitNamingStrategyComponentPathImpl.INSTANCE).build(); sessionFactory = metadata.getSessionFactoryBuilder().build();
session = sessionFactory.getCurrentSession();
transaction = session.beginTransaction();
} @After
public void destory() {
transaction.commit();
session.close();
sessionFactory.close();
}
}

测试:

在TestMain.java中添加测试函数1:

     @Test
public void testInsert() {
Person person = new Person();
person.setName("person1");
person.setAge(27); Student student = new Student();
student.setName("student1");
student.setAge(22);
student.setClassName("class-1");
student.setSchoolName("浙江大学"); session.save(person);
session.save(student);
}

测试执行sql:

 Hibernate: 

     create table PERSONS (
ID integer not null auto_increment,
TYPE varchar(255) not null,
NAME varchar(255),
AGE integer,
CLASS_NAME varchar(255),
SCHOOL_NAME varchar(255),
primary key (ID)
) engine=InnoDB
Hibernate:
insert
into
PERSONS
(NAME, AGE, TYPE)
values
(?, ?, 'person')
Hibernate:
insert
into
PERSONS
(NAME, AGE, CLASS_NAME, SCHOOL_NAME, TYPE)
values
(?, ?, ?, ?, 'student')

数据库查寻结果:

Hibernate(十一):映射继承关系的三种方案

添加测试函数2:

    @Test
public void testSelect() {
List<Person> persons = session.createQuery("FROM Person").list();
System.out.println(persons.size()); List<Student> students = session.createQuery("FROM Student").list();
System.out.println(students.size());
}

执行sql及结果:

 Hibernate:
select
person0_.ID as ID1_0_,
person0_.NAME as NAME3_0_,
person0_.AGE as AGE4_0_,
person0_.CLASS_NAME as CLASS_NA5_0_,
person0_.SCHOOL_NAME as SCHOOL_N6_0_,
person0_.TYPE as TYPE2_0_
from
PERSONS person0_
2
Hibernate:
select
student0_.ID as ID1_0_,
student0_.NAME as NAME3_0_,
student0_.AGE as AGE4_0_,
student0_.CLASS_NAME as CLASS_NA5_0_,
student0_.SCHOOL_NAME as SCHOOL_N6_0_
from
PERSONS student0_
where
student0_.TYPE='student'
1
  • 使用jioned-subclass进行映射

在工程中复制包com.dx.hibernate06.extend,并命名新包名称为:com.dx.hibernate06.joined.subclass

修改Person.hbm.xml:

 <?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated 2017-6-8 15:16:00 by Hibernate Tools 3.5.0.Final -->
<hibernate-mapping package="com.dx.hibernate06.joined.subclass">
<class name="Person" table="PERSONS">
<id name="id" type="java.lang.Integer">
<column name="ID" />
<generator class="native" />
</id> <property name="name" type="java.lang.String">
<column name="NAME" />
</property>
<property name="age" type="java.lang.Integer">
<column name="AGE" />
</property> <joined-subclass name="Student" table="STUDENTS">
<key>
<column name="STUDENT_ID"></column>
</key>
<property name="className" column="CLASS_NAME" type="string"></property>
<property name="schoolName" column="SCHOOL_NAME" type="string"></property>
</joined-subclass>
</class>
</hibernate-mapping>

注意:

1)这里不需要添加discriminator(辨别列)节点配置,也不需要在table和joined-subclass节点中设置discriminator-value(辨别列值);

2)但需要在joined-subclass节点中指定表名,及在节点内部指定key column。

修改hibernate.cfg.xml,修改mapping节点指定文件路径:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
。。。
<mapping resource="com/dx/hibernate06/joined/subclass/Person.hbm.xml" />
</session-factory>
</hibernate-configuration>

测试:

在com.dx.hibernate06.joined.subclass.TestMain.java中执行测试函数testInsert()(备注:该测试函数与上边测试函数一致),执行sql及结果:

 Hibernate: 

     create table PERSONS (
ID integer not null auto_increment,
NAME varchar(255),
AGE integer,
primary key (ID)
) engine=InnoDB
Hibernate: create table STUDENTS (
STUDENT_ID integer not null,
CLASS_NAME varchar(255),
SCHOOL_NAME varchar(255),
primary key (STUDENT_ID)
) engine=InnoDB
Hibernate: alter table STUDENTS
add constraint FK3md9kn7axci4c8qrnaav8ybo
foreign key (STUDENT_ID)
references PERSONS (ID)
Hibernate:
insert
into
PERSONS
(NAME, AGE)
values
(?, ?)
Hibernate:
insert
into
PERSONS
(NAME, AGE)
values
(?, ?)
Hibernate:
insert
into
STUDENTS
(CLASS_NAME, SCHOOL_NAME, STUDENT_ID)
values
(?, ?, ?)

数据库查寻结果:

Hibernate(十一):映射继承关系的三种方案

执行testSelect()测试函数,执行结果及sql:

 Hibernate:
select
person0_.ID as ID1_0_,
person0_.NAME as NAME2_0_,
person0_.AGE as AGE3_0_,
person0_1_.CLASS_NAME as CLASS_NA2_1_,
person0_1_.SCHOOL_NAME as SCHOOL_N3_1_,
case
when person0_1_.STUDENT_ID is not null then 1
when person0_.ID is not null then 0
end as clazz_
from
PERSONS person0_
left outer join
STUDENTS person0_1_
on person0_.ID=person0_1_.STUDENT_ID
2
Hibernate:
select
student0_.STUDENT_ID as ID1_0_,
student0_1_.NAME as NAME2_0_,
student0_1_.AGE as AGE3_0_,
student0_.CLASS_NAME as CLASS_NA2_1_,
student0_.SCHOOL_NAME as SCHOOL_N3_1_
from
STUDENTS student0_
inner join
PERSONS student0_1_
on student0_.STUDENT_ID=student0_1_.ID
1
  • 使用union-subclass进行映射

在工程中复制包com.dx.hibernate06.extend,并命名新包名称为:com.dx.hibernate06.union.subclass

修改Person.hbm.xml:

 <?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!-- Generated 2017-6-8 15:16:00 by Hibernate Tools 3.5.0.Final -->
<hibernate-mapping package="com.dx.hibernate06.union.subclass">
<class name="Person" table="PERSONS">
<id name="id" type="java.lang.Integer">
<column name="ID" />
<generator class="increment" />
</id>
<property name="name" type="java.lang.String">
<column name="NAME" />
</property>
<property name="age" type="java.lang.Integer">
<column name="AGE" />
</property>
<union-subclass name="Student" table="STUDENTS">
<property name="className" column="CLASS_NAME" type="string"></property>
<property name="schoolName" column="SCHOOL_NAME" type="string"></property>
</union-subclass>
</class>
</hibernate-mapping>

注意:

1)这里只需要指定union-subclass对应的表,及指定其独有的列;

2)Person的ID生成类型不能为identity,不能为native,这里采用的increment。

修改hibernate.cfg.xml,修改mapping节点指定文件路径:

 <?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-configuration PUBLIC
"-//Hibernate/Hibernate Configuration DTD 3.0//EN"
"http://www.hibernate.org/dtd/hibernate-configuration-3.0.dtd">
<hibernate-configuration>
<session-factory>
。。。
<mapping resource="com/dx/hibernate06/union/subclass/Person.hbm.xml" />
</session-factory>
</hibernate-configuration>

测试:

在com.dx.hibernate06.union.subclass.TestMain.java中执行测试函数testInsert()(备注:该测试函数与上边测试函数一致),执行sql及结果:

 Hibernate: 

     create table PERSONS (
ID integer not null,
NAME varchar(255),
AGE integer,
primary key (ID)
) engine=InnoDB
Hibernate: create table STUDENTS (
ID integer not null,
NAME varchar(255),
AGE integer,
CLASS_NAME varchar(255),
SCHOOL_NAME varchar(255),
primary key (ID)
) engine=InnoDB
Hibernate:
select
max(ids_.mx)
from
( select
max(ID) as mx
from
STUDENTS
union
select
max(ID) as mx
from
PERSONS
) ids_
Hibernate:
insert
into
PERSONS
(NAME, AGE, ID)
values
(?, ?, ?)
Hibernate:
insert
into
STUDENTS
(NAME, AGE, CLASS_NAME, SCHOOL_NAME, ID)
values
(?, ?, ?, ?, ?)

数据库查寻结果:

Hibernate(十一):映射继承关系的三种方案

执行testSelect()测试函数,执行结果及sql:

 Hibernate:
select
person0_.ID as ID1_0_,
person0_.NAME as NAME2_0_,
person0_.AGE as AGE3_0_,
person0_.CLASS_NAME as CLASS_NA1_1_,
person0_.SCHOOL_NAME as SCHOOL_N2_1_,
person0_.clazz_ as clazz_
from
( select
ID,
NAME,
AGE,
null as CLASS_NAME,
null as SCHOOL_NAME,
0 as clazz_
from
PERSONS
union
select
ID,
NAME,
AGE,
CLASS_NAME,
SCHOOL_NAME,
1 as clazz_
from
STUDENTS
) person0_
2
Hibernate:
select
student0_.ID as ID1_0_,
student0_.NAME as NAME2_0_,
student0_.AGE as AGE3_0_,
student0_.CLASS_NAME as CLASS_NA1_1_,
student0_.SCHOOL_NAME as SCHOOL_N2_1_
from
STUDENTS student0_
1