Create an index on a column of a CollectionTable using Hibernate

Question

Suppose I have the following entity which models a subscriber and uses a CollectionTable to model a list of subscriptions like so:

@Entity
@Table(name = "SUBSCRIBER")
public class Subscriber {
    @ElementCollection
    @CollectionTable(name = "PERSON_ORG_SUBSCRIPTIONS",
                     joinColumns = { @JoinColumn( name = "PERSON_ID", referencedColumnName = "PERSON_ID" ),
                                     @JoinColumn( name = "ORG_ID", referencedColumnName = "ORG_ID" ) } )
    @Column(name = "SUBSCRIPTION_NAME")
    protected Set<String> _subscriptionNames;
}

So this creates a table with columns for PERSON_ID, ORG_ID and SUBSCRIPTION_NAME.

I'm trying to create a database index on the SUBSCRIPTION_NAME column. But if I put the following annotation on _subscriptionNames:

@org.hibernate.annotations.Index( name="subscription_idx", columnNames={"SUBSCRIPTION_NAMES"} )

I get an exception:

org.hibernate.MappingException: Unable to find logical column name from physical name null in table SUBSCRIBER

I also tried using the org.hibernate.annotations.Table annotation on the Subscriber entity, but there does not seem to be a way to have it reference the PERSON_ORG_SUBSCRIPTIONS table.

I'm using Hibernate 3.5.3 and PostgreSQL 9.0.

Have a look here: http://stackoverflow.com/questions/4400169 — Benjamin Seiller, Sep 02 '11 at 09:12
Eventualy I gave up on making some indexes on Hibernate 4.2 and previous. Hibernate 4.3 will support JPA 2.1 which supports `indexes` right inside `@JoinTable`. — Hendy Irawan, Oct 05 '13 at 12:03

score 0 · Answer 1 · answered Dec 30 '11 at 19:22

0

Is the column with name "SUBSCRIPTION_NAME" present in the table SUBSCRIBER?

Are you planning to create index on table from the code? You should propably use hibernate.hbm2ddl.auto=create

answered Dec 30 '11 at 19:22

Create an index on a column of a CollectionTable using Hibernate

1 Answers1