Hi!

I've this mapping:
-----------------------------
<hibernate-mapping package="mm.sistema.model">

<class name="Terceiro" table="terceiros">
<id name="codigo" column="codigo">
<generator class="sequence">
<param name="sequence">ter_seq</param>
</generator>
</id>

<property name="nome" />
<property name="email" />
<property name="fantasia" column="nome_fantasia" />
</class>

<class name="Terceiro$Tipo" table="terceiros_tipos">
<composite-id>
<key-property name="empresa" column="emp_codigo" />
<key-property name="terCodigo" column="ter_codigo" />
<key-property name="tipo" />
</composite-id>

<many-to-one name="terceiro" class="Terceiro" insert="false" update="false" fetch="join" cascade="none">
<column name="ter_codigo" />
</many-to-one>

<property name="dataCadastro" column="data_cadastro" />
<property name="ativo" />
</class>
</hibernate-mapping>
-----------------------------

This HQL returns what I want:
getSession().createQuery(
"select terceiro from Terceiro$Tipo tertip where tertip.terceiro.codigo = 4 and tertip.tipo = 8")
.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY) .list()

How do I write this using Criteria?
Note: I want to return List of Terceiro, not list of Terceiro.Tipo or of array.

Thanks,


Adriano

The where clause seems straightforward. Did you try Projections?

You do something like:

It works, thanks! :-)
But there is still a problem.

It's doing another SELECT to retrieve each item.
The HQL query uses join, I want criteria with joins too.

Can I do this?

Thanks,


Adriano

Add the right setFetchMode("yourRelation",FetchMode.JOIN) on your criteria.

I tried:


and:


And it do:

select * from ( select this_.ter_codigo as y0_ from terceiros_tipos this_ inner join terceiros terceiro1_ on this_.ter_codigo=terceiro1_.codigo ) where rownum <= ?

select terceiro0_.codigo as codigo6_0_, terceiro0_.nome as nome6_0_, terceiro0_.email as email6_0_, terceiro0_.nome_fantasia as nome4_6_0_ from terceiros terceiro0_ where terceiro0_.codigo=?

I'm using Oracle.

Any other tip?

Thanks,


Adriano