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102

answers:

2

According to the Hibernate Reference Documentation it should be possible to mix different inheritance mapping strategies when using Hibernate's XML-Metadata:
http://docs.jboss.org/hibernate/stable/core/reference/en/html/inheritance.html#inheritance-mixing-tableperclass-tablepersubclass

However, the corresponding section of the Hibernate Annotations Reference Guide does not cover that:
http://docs.jboss.org/hibernate/stable/annotations/reference/en/html/entity.html#d0e1168

On the other hand, the JavaDocs suggest that mixing inheritance strategies should be possible. For instance in javax.persistence.DiscriminatorColumn it says:

The strategy and the discriminator column are only specified in the root of an entity class hierarchy or subhierarchy in which a different inheritance strategy is applied.


The following is an example for the mapping I'm trying to achieve. I'd like to use table-per-subclass mapping near the root of the hierarchy, but change to table-per-class-hierarchy mapping near the leaves. Here's some example code:

@Entity
@Inheritance( strategy = InheritanceType.JOINED )
public abstract class A implements Serializable
{
    @Id
    private String id;

    // other mapped properties...
}

@Entity
@Inheritance( strategy = InheritanceType.SINGLE_TABLE )
public class BB extends A
{
    // other mapped properties and associations...
}

@Entity
public class BB1 extends BB
{
    // other stuff, not necessarily mapped...
}

@Entity
public class BB2 extends BB
{
    // other stuff, not necessarily mapped...
}

@Entity
@Inheritance( strategy = InheritanceType.SINGLE_TABLE )
public class CC extends A
{
    // other mapped properties and associations...
}

@Entity
public class CC1 extends CC
{
    // other stuff, not necessarily mapped...
}

...

What I expect from this mapping is having exactly 3 tables: A, BB, and CC. Both BB and CC should have a default discriminator column called DTYPE. They should also provide all columns necessary for all mapped properties and associations of their respective subclasses.

Instead , the class hierarchy seems to use the table-per-subclass inheritance strategy throughout. I.e. I get an own table for each of the entities mentioned above. I'd like to avoid this, since the leaves of the class-hierarchy are extremely light-weight and it just seems overkill to have a separate table for each of them!


Did I overlook something? Any advice is highly appreciated! I'll be glad to provide additional info...

A: 

According to the Hibernate Reference Documentation it should be possible to mix different inheritance mapping strategies when using Hibernate's XML-Metadata (...)

Actually, it's not really supported, they are "cheating" using a secondary table to switch from the single table strategy in the example of the documentation. Quoting Java Persistence with Hibernate:

You can map whole inheritance hierarchies by nesting <union-subclass>, <sub- class>, and <joined-subclass> mapping elements. You can’t mix them — for example, to switch from a table-per-class hierarchy with a discriminator to a normalized table-per-subclass strategy. Once you’ve made a decision for an inheritance strategy, you have to stick to it.

This isn’t completely true, however. With some Hibernate tricks, you can switch the mapping strategy for a particular subclass. For example, you can map a class hierarchy to a single table, but for a particular subclass, switch to a separate table with a foreign key mapping strategy, just as with table per subclass. This is possible with the <join> mapping element:

<hibernate-mapping>
  <class name="BillingDetails"
      table="BILLING_DETAILS">

    <id>...</id>

    <discriminator
        column="BILLING_DETAILS_TYPE"
        type="string"/>
    ...
    <subclass
        name="CreditCard"
        discriminator-value="CC">
      <join table="CREDIT_CARD">
        <key column="CREDIT_CARD_ID"/>

        <property name="number" column="CC_NUMBER"/>
        <property name="expMonth" column="CC_EXP_MONTH"/>
        <property name="expYear" column="CC_EXP_YEAR"/>
        ...
      </join>
    </subclass>

    <subclass
        name="BankAccount"
        discriminator-value="BA">
      <property name=account" column="BA_ACCOUNT"/>
      ...
    </subclass>
  ...
  </class>
</hibernate-mapping>

And you could achieve the same with annotations:

Java Persistence also supports this mixed inheritance mapping strategy with annotations. Map the superclass BillingDetails with InheritanceType.SINGLE_TABLE, as you did before. Now map the subclass you want to break out of the single table to a secondary table.

@Entity
@DiscriminatorValue("CC")
@SecondaryTable(
    name = "CREDIT_CARD",
    pkJoinColumns = @PrimaryKeyJoinColumn(name = "CREDIT_CARD_ID")
)
public class CreditCard extends BillingDetails {
    @Column(table = "CREDIT_CARD",
        name = "CC_NUMBER",
        nullable = false)
    private String number;
    ...
}

I didn't test this but you could maybe try to:

  • map A using a SINGLE_TABLE strategy
  • map BB, CC, etc using the @SecondaryTable annotation.

I've not tested this, I don't know if it will work well for BB1, BB2.

Reference

  • Java Persistence with Hibernate
    • 5.1.5 Mixing inheritance strategies (p207-p210)
Pascal Thivent
Thanks for the quick reply, Pascal. I have to say that the suggested approach seems quite verbose. In particular, I'll have to specify the desired table for each property I map in a subclass. Otherwise it will be mapped to a column of the single table for the hierarchy, right?Well I'll give it a shot tomorrow, and I'll report back...
Meeque
@Meeque Yes, it's verbose. But this is your only option so you don't really have the choice.
Pascal Thivent
Well, this suggestion actually worked well:) However, it still does not feel natural, and there are a few disadvantages.
Meeque
+1  A: 

Just for the sake of clarity, here is Pascal's solution applied to the example code from my question:

@Entity
@Inheritance( strategy = InheritanceType.SINGLE_TABLE )
@DiscriminatorColumn( name = "entityType", 
        discriminatorType = DiscriminatorType.STRING )
public abstract class A implements Serializable
{
    @Id
    private String id;

    // other mapped properties...
}

@Entity
@SecondaryTable( name = "BB" )
public class BB extends A
{
    @Basic( optional = false)
    @Column( table = "BB" )
    private String property1;

    // other mapped properties and associations...
}

@Entity
public class BB1 extends BB
{
    // other stuff, not necessarily mapped...
}

@Entity
public class BB2 extends BB
{
    // other stuff, not necessarily mapped...
}

@Entity
@SecondaryTable( name = "CC" )
public class CC extends A
{
    @ManyToOne( optional = false)
    @JoinColumn( table = "CC" )
    private SomeEntity association1;

    // other mapped properties and associations...
}

@Entity
public class CC1 extends CC
{
    // other stuff, not necessarily mapped...
}

...

I've successfully applied this approach to my problem, and I'll stick to it for the time being. However I still see the following disadvantages:

  • The discriminator column is located in the main table for the hierarchy, the table for root-enity A. In my case, it would be sufficient to have the discriminator column in the secondary tables BB and CC.

  • Anytime one adds properties and associations to subclasses of BB or CC, he/she has to specify that they should be mapped to the respective secondary table. Would be nice, if there was a way to make that the default.

Meeque
+1 for posting the implementation and feedback
Pascal Thivent