CODEWITHSUNDEEP
javainheritancejpaopenjpa
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Reputation: 45

JPA Entity inheritance using DiscriminatorColumn with a 'wild' DiscriminatorValue

I'm trying to implement a inheritence relationship between JPA entities.

Borrowing the example from: http://openjpa.apache.org/builds/1.0.2/apache-openjpa-1.0.2/docs/manual/jpa_overview_mapping_discrim.html

@Entity
@Table(name="SUB", schema="CNTRCT")
@DiscriminatorColumn(name="KIND", discriminatorType=DiscriminatorType.INTEGER)
public abstract class Subscription {
          ...
}

@Entity(name="Lifetime")
@DiscriminatorValue("2")
public class LifetimeSubscription
    extends Subscription {
    ...
}
}

@Entity(name="Trial")
@DiscriminatorValue("3")
public class TrialSubscription
    extends Subscription {
    ...
}

What I need to be able to do is have an additional entity that catches the rest, something like:

  @Entity(name="WildCard")
    @DiscriminatorValue(^[23])
    public class WildSubscription
        extends Subscription {
        ...
    }

Where if it does not match LifetimeSubscription or TrialSubscription it will match WildSubscription. It actually makes a bit more sense if you think of it where the wild is the superclass, and if there is not a more concrete implementation that fits, use the superclass.

Anyone know of a method of doing this?

Thanks!

Upvotes: 1

Views: 1029

Answers (1)

MaDa
MaDa

Reputation: 10762

The JPA API allows only plain values here, and for a reason: discriminator values are mapped to SQL WHERE:

SELECT ... WHERE kind = 1

If you could specify regular expressions here, it wouldn't be transferable to SQL, as it does not support such constructs.

Upvotes: 1

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