Hibernate 作为ORM，面对的一个问题是：一个方面是关系型的数据库，另一面是Java Object。Java作为一种面向对象的语言支持继承关系。Hibernate本身提供了4种策略将继承关系映射到关系型的数据库中。他们分别是:
- Table Per Concrete Class with Implicit Polymorphism : 每个Concrete Class 对应一张Table。利用Hibernate内部的机制来实现多态型的查询。对于SuperClass Query是分成对Concrete Class的独立Query组合起来的。
- Table Per Concrete Class with Union : 每个Concrete Class 对应一张Table。对于SuperClass Query是通过Union的方式来实现的。
- Table Per Class Hierarchy : 将整个Class Hierachy的关系Mapping到一张Table中，通过discriminator来区分具体的Concrete Class。
- Table Per SubClass:对于Class Hierachy图中的每个类都对应有一张Table。
这些策略各有特点，应用场合不同。在具体实现的时候，应该仔细选择。
一、例子的背景：
假如我们有如下的类的关系，需要将他们mapping到数据库中
每个类的代码如下：
package org.example.hibernate.domain;
//BillDetail SuperClass
public class BillingDetail {
private Long id;
private String owner;
//省略相应的get/set accessor
}
package org.example.hibernate.domain;
//Concrete Class BankAccount
public class BankAccount extends BillingDetail{
private String account;
private String bankName;
private String swift;
//省略相应的get/set accessor
}
package org.example.hibernate.domain;
//Concrete Class CreditCard
public class CreditCard extends BillingDetail {
private String number;
private String expMonth;
private String expYear;
//省略相应的get/set accessor
}
二、使用T/CCIP(Table Per Concrete Class with Implicit Polymorphism)策略来映射
1. 只对Concrete Class建立Table，SuperClass中的property，直接mapping到Concrete Class相应Table的column
2. 对于Concrete Class的mapping，就如同没有SuperClass的普通Class一样mapping。
Mapping File如下：
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE hibernate-mapping PUBLIC
"-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
<!-- CreditCard Class Mapping Section-->
<class name="org.example.hibernate.domain.CreditCard"
table="CREDITCARD" polymorphism="implicit">
<id name="id" column="IDNUM" type="long">
<generator class="native"/>
</id>
<property name="owner" type="string" column="OWNER" length="30"/>
<property name="number" type="string" column="NUMBER_CODE" length="20"/>
<property name="expMonth" type="string" column="EXP_MONTH" length="2"/>
<property name="expYear" type="string" column="EXP_YEAR" length="4"/>
</class>
<!-- BankAccount Class Mapping Section-->
<class name="org.example.hibernate.domain.BankAccount"
table="BANK_ACCOUNT" polymorphism="implicit">
<id name="id" column="IDNUM" type="long">
<generator class="native"/>
</id>
<property name="owner" type="string" column="OWNER" length="30"/>
<property name="account" type="string" column="ACCOUNT" length="20"/>
<property name="bankName" type="string" column="BANK_NAME" length="255"/>
<property name="swift" type="string" column="SWIFT" length="3"/>
</class>
</hibernate-mapping>
如Mapping File中看到的，CreditCard和BankAccount的Id都是采用Hibernate的native策略。当Session对CreditCard和BankAccount的instance做save的时候，insert到相应table中的idnum栏位的value可能是一样的。因为他们的Hibernate Id 产生策略独立的，彼此没有任何的关系。
//Transaction Commit以后，cc11的Id Value是1
CreditCard cc11 = new CreditCard();
cc11.setOwner("jessecia");
cc11.setNumber("123123123123123");
session.save(cc11);
//Transaction Commit以后，ba11的Id Value是1
BankAccount ba11 = new BankAccount();
ba11.setOwner("jessecia");
ba11.setBankName("GBDFB");
session.save(ba11);
而对于BillingDetail的polymorphic query，则分成2个独立的对CreditCard和BankAccount表的查询SQL来获得。Java代码如下：
Criteria criteria = session.createCriteria(BillingDetail.class);
List<BillingDetail> list = criteria.list();
产生的SQL如下：
select this_.IDNUM as IDNUM0_0_
, this_.NUMBER_CODE as NUMBER2_0_0_
, this_.EXP_MONTH as EXP3_0_0_
, this_.EXP_YEAR as EXP4_0_0_
, this_.OWNER as OWNER0_0_
from CREDITCARD this_
select this_.IDNUM as IDNUM1_0_
, this_.ACCOUNT as ACCOUNT1_0_
, this_.BANK_NAME as BANK3_1_0_
, this_.SWIFT as SWIFT1_0_
, this_.OWNER as OWNER1_0_
from BANK_ACCOUNT1 this_
T/CCIP方式的结论：
1. 应用场合：不适合于Polymorphic Association的应用场合。也就是说SupperClass和其他的Persistent Class发生关系。
2. 扩展性：当SuperClass增加新的Property的时候，每个Concrete Class相应的Table都需要新增Column
