第一种 一个员工类它有两个子类,利用hibernate我们把这员工类及子类都映射到一张表里,员工表是普通员工,子类的 话是技术员工的类和销售员工 那么这三个类怎么映射到一张表里,我们又如何区分这员工的到底是哪一类员工呢 这时,我们一般在表中加入了一个识别的字段,比如说type 这个字段有三个值,默认值为0 表示的普通员工,1是 技术员工 2是销售员工 好,现在开始 设计主体员工实体类 package vo.util.bean; /** * 员工类 * @author Administrator * */ public class Employee { private int id; private String name; private Department derpartment; public int getId() { return id; } public void setId(int id) { this.id = id; } public String getName() { return name; } public void setName(String name) { this.name = name; } public Department getDerpartment() { return derpartment; } public void setDerpartment(Department derpartment) { this.derpartment = derpartment; } } 技术员工类 继承主体员工类 package vo.util.bean; public class Skiller extends Employee { private String skill; public String getSkill() { return skill; } public void setSkill(String skill) { this.skill = skill; } } 销售员工类 继承主体员工类 package vo.util.bean; public class Sales extends Employee { private int sale; public int getSale() { return sale; } public void setSale(int sale) { this.sale = sale; } } 主体实体映射文件 <?xml version="1.0"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> <hibernate-mapping package="vo.util.bean"> <class name="Employee" discriminator-value="0"> <id name="id"> <generator class="native"/> </id> <discriminator column="type"/> <property name="name"/> <many-to-one name="derpartment" column="depaer_id" not-null="true"/> <subclass name="Sales" discriminator-value="1"> <property name="sale"/> </subclass> <subclass name="Skiller" discriminator-value="2"> <property name="skill"/> </subclass> </class> </hibernate-mapping> 分析 subclass 就是子类的意思 在这里的节点就是配置就是主实体的子类的配置 name 子实体名称 property 子实体的属性 discriminator-value识别字段值 discriminator节点 加入主表的识别字段 column="type"就是识别字段的列名是type 这个列默认的字段默认类型是string 可以加入 type="string" 就可以设计这个列的值的 类型 测试类 package vo.util.test; import java.util.HashSet; import java.util.Set; import org.hibernate.Session; import org.hibernate.Transaction; import vo.util.HibernateUtil; import vo.util.bean.Department; import vo.util.bean.Employee; import vo.util.bean.Sales; import vo.util.bean.Skiller; public class OneToMany { /** * @param args */ public static void main(String[] args) { add(); query(1); } static Department add(){ Session session = null; Transaction tx = null; try{ session =HibernateUtil.getSession(); tx = session.beginTransaction(); //…你的代码save,delete,update,get… Department dpart=new Department(); dpart.setName("政企事业部"); Employee em1=new Employee(); em1.setName("员工李小仙"); em1.setDerpartment(dpart); Employee em2=new Employee(); em2.setName("员工张小三"); em2.setDerpartment(dpart); Skiller s=new Skiller(); s.setDerpartment(dpart); s.setName("员工高明明"); s.setSkill("skill"); Sales sl=new Sales(); sl.setDerpartment(dpart); sl.setName("小凤仙"); sl.setSale(2); /** Set<Employee> em=new HashSet<Employee>(); em.add(em1); em.add(em2); em.add(sl); em.add(s); dpart.setEms(em); */ session.save(dpart); session.save(em1); session.save(em2); session.save(s); session.save(sl); tx.commit(); return dpart; }finally{ if(session != null)session.close(); } } static Department query(int id){ Session session=null; try{ session=HibernateUtil.getSession(); Department d=(Department)session.get(Department.class, id); System.out.println(d.getName()+"--------->"+d.getEms().getClass().getName()); return d; }finally{ if(session != null)session.close(); } } } 映射的表的值 1,"0","员工李小仙",1,NULL,NULL 2,"0","员工张小三",1,NULL,NULL 3,"2","员工高明明",1,NULL,"skill" 4,"1","小凤仙", 1, 2, NULL 发现表中的值有很多值为空值 现在我们来看表结构 CREATE TABLE `employee` ( `id` int(11) NOT NULL auto_increment, `type` varchar(255) NOT NULL, `name` varchar(255) default NULL, `depaer_id` int(11) NOT NULL, `sale` int(11) default NULL, `skill` varchar(255) default NULL, PRIMARY KEY (`id`), KEY `FK4AFD4ACE59BDCD94` (`depaer_id`), CONSTRAINT `FK4AFD4ACE59BDCD94` FOREIGN KEY (`depaer_id`) REFERENCES `department` (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; 但是这样是不是不合理呢?怎么修正它呢 ? 第二种方式 可以这样来,我们可以根据子类实体也新建一张表,比如技术员工,就新建一张技术员工表 同时新建的表如何跟主表employee连接呢?我们可以这样类设计,子表的id就是employee的id 这个字表的id不但是主键同时也是一个外键,引用的是主表的id 那现在我们来设计吧 joined-subclass节点的作用就出来了 主实体和子实体都不变 只需要修改员工的employee的映射文件便可 <?xml version="1.0"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> <hibernate-mapping package="vo.util.bean"> <class name="Employee"> <id name="id"> <generator class="native"/> </id> <!-- <discriminator column="type"/>--> <property name="name"/> <many-to-one name="derpartment" column="depaer_id" not-null="true"/> <!-- <subclass name="Sales" discriminator-value="1"> <property name="sale"/> </subclass> <subclass name="Skiller" discriminator-value="2"> <property name="skill"/> </subclass> --> <joined-subclass name="Sales"> <key column="emp_id"/> <property name="sale"/> </joined-subclass> <joined-subclass name="Skiller"> <key column="emp_id"/> <property name="skill"/> </joined-subclass> </class> </hibernate-mapping> 分析:joined-subclass 节点就是引入子实体类 key节点就是外键主键名称 property 就是子实体的属性 测试类不变 测试后输出的sql log4j:WARN No appenders could be found for logger (org.hibernate.cfg.Environment). log4j:WARN Please initialize the log4j system properly. 打开连接成功!Hibernate: insert into Department (name) values (?) Hibernate: insert into Employee (name, depaer_id) values (?, ?) Hibernate: insert into Employee (name, depaer_id) values (?, ?) Hibernate: insert into Employee (name, depaer_id) values (?, ?) Hibernate: insert into Skiller (skill, emp_id) values (?, ?) Hibernate: insert into Employee (name, depaer_id) values (?, ?) Hibernate: insert into Sales (sale, emp_id) values (?, ?) Hibernate: select department0_.id as id0_0_, department0_.name as name0_0_ from Department department0_ where department0_.id=? Hibernate: select ems0_.depaer_id as depaer3_1_, ems0_.id as id1_, ems0_.id as id1_0_, ems0_.name as name1_0_, ems0_.depaer_id as depaer3_1_0_, ems0_1_.sale as sale2_0_, ems0_2_.skill as skill3_0_, case when ems0_1_.emp_id is not null then 1 when ems0_2_.emp_id is not null then 2 when ems0_.id is not null then 0 end as clazz_0_ from Employee ems0_ left outer join Sales ems0_1_ on ems0_.id=ems0_1_.emp_id left outer join Skiller ems0_2_ on ems0_.id=ems0_2_.emp_id where ems0_.depaer_id=? 政企事业部--------->[vo.util.bean.Employee@fcfa52, vo.util.bean.Sales@1fc6e42, vo.util.bean.Skiller@18b81e3, vo.util.bean.Employee@961dff] 现在来看表结构 CREATE TABLE `employee` ( `id` int(11) NOT NULL auto_increment, `name` varchar(255) default NULL, `depaer_id` int(11) NOT NULL, PRIMARY KEY (`id`), KEY `FK4AFD4ACE59BDCD94` (`depaer_id`), CONSTRAINT `FK4AFD4ACE59BDCD94` FOREIGN KEY (`depaer_id`) REFERENCES `department` (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; 子表 CREATE TABLE `sales` ( `emp_id` int(11) NOT NULL, `sale` int(11) default NULL, PRIMARY KEY (`emp_id`), KEY `FK4BF58ECF47ACFA7` (`emp_id`), CONSTRAINT `FK4BF58ECF47ACFA7` FOREIGN KEY (`emp_id`) REFERENCES `employee` (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; CREATE TABLE `skiller` ( `emp_id` int(11) NOT NULL, `skill` varchar(255) default NULL, PRIMARY KEY (`emp_id`), KEY `FKE3361E9EF47ACFA7` (`emp_id`), CONSTRAINT `FKE3361E9EF47ACFA7` FOREIGN KEY (`emp_id`) REFERENCES `employee` (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8; 第三种方式 为了效率的提高 我们可以结合这两种的映射配置映射文件,可以去试下吧 第四种方式 员工是一张完整的表 技术员工 销售员工都是一张完整的信息表 主实体及子实体都不变 修改Employee的映射文件 <?xml version="1.0"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> <hibernate-mapping package="vo.util.bean"> <class name="Employee" discriminator-value="0"> <id name="id"> <generator class="native"/> </id> <!-- <discriminator column="type"/> --> <property name="name"/> <many-to-one name="derpartment" column="depaer_id" not-null="true"/> <!-- <subclass name="Sales" discriminator-value="1"> <property name="sale"/> </subclass> <subclass name="Skiller" discriminator-value="2"> <property name="skill"/> </subclass> --> <union-subclass name="Sales" table="tables"> <property name="sale"/> </union-subclass> <union-subclass name="Skiller" table="skill"> <property name="skill"/> </union-subclass> </class> </hibernate-mapping> union-subclass节点配置 name="Sales" 就是对应的实体类 table 给子实体指定的表 property 节点是属性名 就是对应子实体的属性 测试类不变 运行测试 log4j:WARN No appenders could be found for logger (org.hibernate.cfg.Environment). log4j:WARN Please initialize the log4j system properly. Exception in thread "main" java.lang.ExceptionInInitializerError at vo.util.test.OneToMany.add(OneToMany.java:29) at vo.util.test.OneToMany.main(OneToMany.java:22) Caused by: org.hibernate.MappingException: Cannot use identity column key generation with <union-subclass> mapping for: vo.util.bean.Employee at org.hibernate.persister.entity.UnionSubclassEntityPersister.<init>(UnionSubclassEntityPersister.java:67) at org.hibernate.persister.PersisterFactory.createClassPersister(PersisterFactory.java:61) at org.hibernate.impl.SessionFactoryImpl.<init>(SessionFactoryImpl.java:226) at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1294) at vo.util.HibernateUtil.<clinit>(HibernateUtil.java:23) ... 2 more 发现报实体异常,意思说说主键的生成器 identity 问题 这是什么原因造成的呢? 主键生成错误 看配置文件 <generator class="native"/> 这种主键生成是自增长得,那么这个三张表会产生 主键值是一样的 主键值代表的是一条记录 所以可以改成hilo end 完毕!