9장. Inheritance mapping

It is possible to use different mapping strategies for different branches of the same inheritance hierarchy. You can then make use of implicit polymorphism to achieve polymorphism across the whole hierarchy. However, Hibernate does not support mixing <subclass>, <joined-subclass> and <union-subclass> mappings under the same root <class> element. It is possible to mix together the table per hierarchy and table per subclass strategies under the the same <class> element, by combining the <subclass> and <join> elements (see below for an example).

It is possible to define subclass, union-subclass, and joined-subclass mappings in separate mapping documents directly beneath hibernate-mapping. This allows you to extend a class hierarchy by adding a new mapping file. You must specify an extends attribute in the subclass mapping, naming a previously mapped superclass. Previously this feature made the ordering of the mapping documents important. Since Hibernate3, the ordering of mapping files is irrelevant when using the extends keyword. The ordering inside a single mapping file still needs to be defined as superclasses before subclasses.

 <hibernate-mapping>
     <subclass name="DomesticCat" extends="Cat" discriminator-value="D">
          <property name="name" type="string"/>
     </subclass>
 </hibernate-mapping>

9.1.1. Table per class hierarchy

Suppose we have an interface Payment with the implementors CreditCardPayment, CashPayment, and ChequePayment. The table per hierarchy mapping would display in the following way:

<class name="Payment" table="PAYMENT">
    <id name="id" type="long" column="PAYMENT_ID">
        <generator class="native"/>
    </id>
    <discriminator column="PAYMENT_TYPE" type="string"/>
    <property name="amount" column="AMOUNT"/>
    ...
    <subclass name="CreditCardPayment" discriminator-value="CREDIT">
        <property name="creditCardType" column="CCTYPE"/>
        ...
    </subclass>
    <subclass name="CashPayment" discriminator-value="CASH">
        ...
    </subclass>
    <subclass name="ChequePayment" discriminator-value="CHEQUE">
        ...
    </subclass>
</class>

Exactly one table is required. There is a limitation of this mapping strategy: columns declared by the subclasses, such as CCTYPE, cannot have NOT NULL constraints.

9.1.2. Table per subclass

A table per subclass mapping looks like this:

<class name="Payment" table="PAYMENT">
    <id name="id" type="long" column="PAYMENT_ID">
        <generator class="native"/>
    </id>
    <property name="amount" column="AMOUNT"/>
    ...
    <joined-subclass name="CreditCardPayment" table="CREDIT_PAYMENT">
        <key column="PAYMENT_ID"/>
        <property name="creditCardType" column="CCTYPE"/>
        ...
    </joined-subclass>
    <joined-subclass name="CashPayment" table="CASH_PAYMENT">
        <key column="PAYMENT_ID"/>
        ...
    </joined-subclass>
    <joined-subclass name="ChequePayment" table="CHEQUE_PAYMENT">
        <key column="PAYMENT_ID"/>
        ...
    </joined-subclass>
</class>

Four tables are required. The three subclass tables have primary key associations to the superclass table so the relational model is actually a one-to-one association.

9.1.3. Table per subclass: using a discriminator

Hibernate's implementation of table per subclass does not require a discriminator column. Other object/relational mappers use a different implementation of table per subclass that requires a type discriminator column in the superclass table. The approach taken by Hibernate is much more difficult to implement, but arguably more correct from a relational point of view. If you want to use a discriminator column with the table per subclass strategy, you can combine the use of <subclass> and <join>, as follows:

<class name="Payment" table="PAYMENT">
    <id name="id" type="long" column="PAYMENT_ID">
        <generator class="native"/>
    </id>
    <discriminator column="PAYMENT_TYPE" type="string"/>
    <property name="amount" column="AMOUNT"/>
    ...
    <subclass name="CreditCardPayment" discriminator-value="CREDIT">
        <join table="CREDIT_PAYMENT">
            <key column="PAYMENT_ID"/>
            <property name="creditCardType" column="CCTYPE"/>
            ...
        </join>
    </subclass>
    <subclass name="CashPayment" discriminator-value="CASH">
        <join table="CASH_PAYMENT">
            <key column="PAYMENT_ID"/>
            ...
        </join>
    </subclass>
    <subclass name="ChequePayment" discriminator-value="CHEQUE">
        <join table="CHEQUE_PAYMENT" fetch="select">
            <key column="PAYMENT_ID"/>
            ...
        </join>
    </subclass>
</class>

선택적인 fetch="select" 선언은 슈퍼클래스를 질의할 때 outer join을 사용하여 ChequePayment 서브클래스 데이터를 페치시키지 않도록 Hibernate에게 알려준다.

9.1.4. table per class hierarchy와 table per subclass를 혼합하기

You can even mix the table per hierarchy and table per subclass strategies using the following approach:

<class name="Payment" table="PAYMENT">
    <id name="id" type="long" column="PAYMENT_ID">
        <generator class="native"/>
    </id>
    <discriminator column="PAYMENT_TYPE" type="string"/>
    <property name="amount" column="AMOUNT"/>
    ...
    <subclass name="CreditCardPayment" discriminator-value="CREDIT">
        <join table="CREDIT_PAYMENT">
            <property name="creditCardType" column="CCTYPE"/>
            ...
        </join>
    </subclass>
    <subclass name="CashPayment" discriminator-value="CASH">
        ...
    </subclass>
    <subclass name="ChequePayment" discriminator-value="CHEQUE">
        ...
    </subclass>
</class>

이들 매핑 방도들 중 어떤 것에 대해, 루트 Payment 클래스에 대한 하나의 다형성 연관은 <many-to-one>을 사용하여 매핑된다.

<many-to-one name="payment" column="PAYMENT_ID" class="Payment"/>

9.1.5. Table per concrete class

There are two ways we can map the table per concrete class strategy. First, you can use <union-subclass>.

<class name="Payment">
    <id name="id" type="long" column="PAYMENT_ID">
        <generator class="sequence"/>
    </id>
    <property name="amount" column="AMOUNT"/>
    ...
    <union-subclass name="CreditCardPayment" table="CREDIT_PAYMENT">
        <property name="creditCardType" column="CCTYPE"/>
        ...
    </union-subclass>
    <union-subclass name="CashPayment" table="CASH_PAYMENT">
        ...
    </union-subclass>
    <union-subclass name="ChequePayment" table="CHEQUE_PAYMENT">
        ...
    </union-subclass>
</class>

세 개의 테이블들이 슈퍼클래스들에 대해 수반된다. 각각의 테이블은 상속된 프로퍼티들을 포함하여, 그 클래스의 모든 프로퍼티들에 대한 컬럼들을 정의한다.

The limitation of this approach is that if a property is mapped on the superclass, the column name must be the same on all subclass tables. The identity generator strategy is not allowed in union subclass inheritance. The primary key seed has to be shared across all unioned subclasses of a hierarchy.

If your superclass is abstract, map it with abstract="true". If it is not abstract, an additional table (it defaults to PAYMENT in the example above), is needed to hold instances of the superclass.

9.1.6. Table per concrete class using implicit polymorphism

대안적인 접근법은 함축적인 다형성을 사용하는 것이다:

<class name="CreditCardPayment" table="CREDIT_PAYMENT">
    <id name="id" type="long" column="CREDIT_PAYMENT_ID">
        <generator class="native"/>
    </id>
    <property name="amount" column="CREDIT_AMOUNT"/>
    ...
</class>

<class name="CashPayment" table="CASH_PAYMENT">
    <id name="id" type="long" column="CASH_PAYMENT_ID">
        <generator class="native"/>
    </id>
    <property name="amount" column="CASH_AMOUNT"/>
    ...
</class>

<class name="ChequePayment" table="CHEQUE_PAYMENT">
    <id name="id" type="long" column="CHEQUE_PAYMENT_ID">
        <generator class="native"/>
    </id>
    <property name="amount" column="CHEQUE_AMOUNT"/>
    ...
</class>

Notice that the Payment interface is not mentioned explicitly. Also notice that properties of Payment are mapped in each of the subclasses. If you want to avoid duplication, consider using XML entities (for example, [ <!ENTITY allproperties SYSTEM "allproperties.xml"> ] in the DOCTYPE declaration and &allproperties; in the mapping).

이 접근법의 단점은 다형성 질의들을 수행할 때 Hibernate가 생성된 SQl UNION들을 생성시키는 않는다는 점이다.

이 매핑 방도의 경우, Payment에 대한 하나의 다형성 연관은 대개 <any>를 사용하여 매핑된다.

<any name="payment" meta-type="string" id-type="long">
    <meta-value value="CREDIT" class="CreditCardPayment"/>
    <meta-value value="CASH" class="CashPayment"/>
    <meta-value value="CHEQUE" class="ChequePayment"/>
    <column name="PAYMENT_CLASS"/>
    <column name="PAYMENT_ID"/>
</any>

9.1.7. 함축적인 다형성을 다른 상속 매핑들과 혼합하기

Since the subclasses are each mapped in their own <class> element, and since Payment is just an interface), each of the subclasses could easily be part of another inheritance hierarchy. You can still use polymorphic queries against the Payment interface.

<class name="CreditCardPayment" table="CREDIT_PAYMENT">
    <id name="id" type="long" column="CREDIT_PAYMENT_ID">
        <generator class="native"/>
    </id>
    <discriminator column="CREDIT_CARD" type="string"/>
    <property name="amount" column="CREDIT_AMOUNT"/>
    ...
    <subclass name="MasterCardPayment" discriminator-value="MDC"/>
    <subclass name="VisaPayment" discriminator-value="VISA"/>
</class>

<class name="NonelectronicTransaction" table="NONELECTRONIC_TXN">
    <id name="id" type="long" column="TXN_ID">
        <generator class="native"/>
    </id>
    ...
    <joined-subclass name="CashPayment" table="CASH_PAYMENT">
        <key column="PAYMENT_ID"/>
        <property name="amount" column="CASH_AMOUNT"/>
        ...
    </joined-subclass>
    <joined-subclass name="ChequePayment" table="CHEQUE_PAYMENT">
        <key column="PAYMENT_ID"/>
        <property name="amount" column="CHEQUE_AMOUNT"/>
        ...
    </joined-subclass>
</class>

Once again, Payment is not mentioned explicitly. If we execute a query against the Payment interface, for example from Payment, Hibernate automatically returns instances of CreditCardPayment (and its subclasses, since they also implement Payment), CashPayment and ChequePayment, but not instances of NonelectronicTransaction.

9.2. 제약들

There are limitations to the "implicit polymorphism" approach to the table per concrete-class mapping strategy. There are somewhat less restrictive limitations to <union-subclass> mappings.

다음 표는 Hibernate에서 table per concrete-class 매핑들에 대한 제약들, 그리고 함축적인 다형성에 대한 제약들을 보여준다.

표 9.1. 상속 매핑들의 특징들

상속 방도	다형성 다대일	다형성 일대일	다형성 일대다	다형성 다대다	Polymorphic `load()/get()`	다형성 질의들	다형성 조인들	Outer 조인 페칭
table per class-hierarchy	`<many-to-one>`	`<one-to-one>`	`<one-to-many>`	`<many-to-many>`	`s.get(Payment.class, id)`	`from Payment p`	`from Order o join o.payment p`	지원됨
table per subclass	`<many-to-one>`	`<one-to-one>`	`<one-to-many>`	`<many-to-many>`	`s.get(Payment.class, id)`	`from Payment p`	`from Order o join o.payment p`	지원됨
table per concrete-class (union-subclass)	`<many-to-one>`	`<one-to-one>`	`<one-to-many>` (for `inverse="true"` only)	`<many-to-many>`	`s.get(Payment.class, id)`	`from Payment p`	`from Order o join o.payment p`	지원됨
table per concrete class (implicit polymorphism)	`<any>`	지원되지 않음	지원되지 않음	`<many-to-any>`	`s.createCriteria(Payment.class).add( Restrictions.idEq(id) ).uniqueResult()`	`from Payment p`	지원되지 않음	지원되지 않음