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Hibernate version: 3.0.5

Mapping documents:

TestataProposta.hbm.xml :

<hibernate-mapping>
<class name="it.xxx.scm.TestataProposta" table="testata_proposta">

<id name="id" column="id" type="long">
<generator class="sequence" />
</id>

<!-- Business key -->
<property name="idCommittente" type="string" column="id_committente" not-null="true"/>
<property name="idProposta" type="string" column="id_proposta"/>
<!-- Business key -->

<property name="sezione" type="string" column="sezione"/>
<property name="tipoOrdine" type="string" column="tipo_ordine" />
<property name="data" type="java.util.Date" column="data" />
<property name="destinatario" type="string" column="destinatario" />

<set name="righeProposta" lazy="true" inverse="true" cascade="all" outer-join="true">
<key>
<column name="id_proposta" />
</key>
<one-to-many class="it.xxx.scm.RigaProposta" />
</set>

</class>
</hibernate-mapping>

RigaProposta.hbm.xml :

<hibernate-mapping>
<class name="it.xxx.scm.RigaProposta" table="riga_proposta">

<id name="id" column="id" type="long">
<generator class="sequence" />
</id>

<!-- Business key -->
<property name="idCommittente" type="string" column="id_committente" not-null="true"/>
<property name="idProposta" type="string" column="id_proposta" insert="false" update="false" />
<property name="riga" type="java.lang.Integer" column="riga" />
<!-- Business key -->

<property name="idArticolo" type="string" column="id_articolo" />
<property name="quantita" type="java.math.BigDecimal" column="quantita" />

<many-to-one name="testataProposta" class="it.xxx.scm.TestataProposta" property-ref="idProposta">
<column name="id_proposta" />
</many-to-one>

</class>
</hibernate-mapping>


Code between sessionFactory.openSession() and session.close():

TestataProposta testata = session.get(TestataProposta.class, id);

Full stack trace of any exception that occurs: None


Name and version of the database you are using: PostgreSQL 7.4

The generated SQL (show_sql=true):

Hibernate: select testatapro0_.id as id1_, testatapro0_.id_committente as id2_40_1_, testatapro0_.id_proposta as id3_40_1_, test
atapro0_.sezione as sezione40_1_, testatapro0_.tipo_ordine as tipo5_40_1_, testatapro0_.data as data40_1_, testatapro0_.destinat
ario as destinat7_40_1_, righepropo1_.id_proposta as id3___, righepropo1_.id as id__, righepropo1_.id as id0_, righepropo1_.id_c
ommittente as id2_41_0_, righepropo1_.id_proposta as id3_41_0_, righepropo1_.riga as riga41_0_, righepropo1_.id_articolo as id5_
41_0_, righepropo1_.quantita as quantita41_0_ from testata_proposta testatapro0_ left outer join riga_proposta righepropo1_ on t
estatapro0_.id=righepropo1_.id_proposta where testatapro0_.id=?

Debug level Hibernate log excerpt: None useful


I'm wondering if what I'm trying to do is possible.

I have a parent table, TestataProposta, and a child table, RigaProposta.

They both have their own auto-generated primary key id. TestataProposta has a business (natural let's say) key, called "id_proposta". The children have that column too, so they can <many-to-one> to the father (and this works).

When I wrote like 150 TestataProposta object with children Hibernate correctly wrote to the database the parent and children objects with cascade="all" (so I just called testataProposta.save()), and populated the "id_proposta" column on children objects with the correct business value (not the primary key).

But when I do a query on TestataProposta, Hibernate tries to match parent's primary key (ID) with children business key (ID_PROPOSTA). And of course the parent has no children, this way. See the generated SQL above : on testatapro0_.id=righepropo1_.id_proposta

I did not find any way to specify on the <one-to-many> clause the source column for the association. Is this impossibile ?

I don't want to use the primary key for the association, because if we export the data to another database that key would be probably meaningless.

Any suggestion ?

Thanks

Giulio