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Hibernateは3つの基本的な継承のマッピング戦略をサポートします。
クラス階層ごとのテーブル（table-per-class-hierarchy）
table per subclass
具象クラスごとのテーブル（table-per-concrete-class）
加えて4つ目に、Hibernateはわずかに異なる性質を持ったポリモーフィズムをサポートします。
暗黙的ポリモーフィズム
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>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.
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.
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>オプションの fetch="select" 宣言は、 スーパークラスのクエリ実行時に外部結合を使って、 サブクラスの ChequePayment データを取得しないように指定するためのものです。
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>いずれのマッピング戦略であっても、ルートである Payment クラスへの ポリモーフィックな関連は <many-to-one> を使ってマッピングします。
<many-to-one name="payment" column="PAYMENT_ID" class="Payment"/>
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>サブクラスごとに3つのテーブルが必要です。それぞれのテーブルは、継承プロパティを含んだ、 クラスの全てのプロパティに対するカラムを定義します。
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.
もう一つのアプローチは暗黙的ポリモーフィズムの使用です。
<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).
このアプローチの欠点は、Hibernateがポリモーフィックなクエリの実行時にSQL UNION を生成しない点です。
このマッピング戦略に対しては、Payment へのポリモーフィックな関連は 通常、<any> を使ってマッピングされます。
<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>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.
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.
次のテーブルに、Hibernateにおけるtable-per-concrete-classマッピングの 制限や暗黙的ポリモーフィズムの制限を示します。
表 9.1. 継承マッピングの機能
| 継承戦略 | 多対一のポリモーフィズム | 一対一のポリモーフィズム | 一対多のポリモーフィズム | 多対多のポリモーフィズム | Polymorphic load()/get() | ポリモーフィズムを使ったクエリ | ポリモーフィズムを使った結合 | 外部結合によるフェッチ |
|---|---|---|---|---|---|---|---|---|
| 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 | サポート |
| 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 | サポート |
| 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 | サポート |
| table per concrete class (implicit polymorphism) | <any> | サポートしていません | サポートしていません | <many-to-any> | s.createCriteria(Payment.class).add( Restrictions.idEq(id) ).uniqueResult() | from Payment p | サポートしていません | サポートしていません |
製作著作 © 2004 Red Hat Middleware, LLC.