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Hibernate supports the three basic inheritance mapping strategies:
table per class hierarchy
table per subclass
table per concrete class
In addition, Hibernate supports a fourth, slightly different kind of polymorphism:
implicit polymorphism
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 Hibernate, 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>
The optional fetch="select"
declaration tells Hibernate
not to fetch the ChequePayment
subclass data using an
outer join when querying the superclass.
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>
For any of these mapping strategies, a polymorphic association to the root
Payment
class is mapped using
<many-to-one>
.
<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>
Three tables are involved for the subclasses. Each table defines columns for all properties of the class, including inherited properties.
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.
An alternative approach is to make use of implicit polymorphism:
<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).
The disadvantage of this approach is that Hibernate does not generate SQL
UNION
s when performing polymorphic queries.
For this mapping strategy, a polymorphic association to Payment
is usually mapped using <any>
.
<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.
The following table shows the limitations of table per concrete-class mappings, and of implicit polymorphism, in Hibernate.
Table 10.1. Features of inheritance mappings
Inheritance strategy | Polymorphic many-to-one | Polymorphic one-to-one | Polymorphic one-to-many | Polymorphic many-to-many | Polymorphic load()/get() | Polymorphic queries | Polymorphic joins | Outer join fetching |
---|---|---|---|---|---|---|---|---|
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 | supported |
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 | supported |
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 | supported |
table per concrete class (implicit polymorphism) | <any> | not supported | not supported | <many-to-any> | s.createCriteria(Payment.class).add( Restrictions.idEq(id) ).uniqueResult() | from Payment p | not supported | not supported |