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Forum locked This topic is locked, you cannot edit posts or make further replies.  [ 3 posts ] 
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 Post subject: CreateFilter problem...
PostPosted: Fri Dec 16, 2005 7:19 am 
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Joined: Wed Sep 22, 2004 8:27 am
Posts: 89
Hibernate version: 3.1

Mapping documents:
Code:
<hibernate-mapping default-lazy="false">

    <class name="com.coginfo.kf.bean.Categoria" table="KFT_CATEGORIA" select-before-update="true">
        <id name="codint" type="long" unsaved-value="0">
            <generator class="sequence">
               <param name="sequence">KF_SEQ_CATEGORIA</param>
            </generator>   
         </id>

        <property name="cafdsc" type="string">
            <column name="cafdsc" length="1000" sql-type="string" not-null="true"/>
        </property>


      <set name="categoriaRuoli"  lazy="true" cascade="none">
         <key column="codintctg"/>
         <one-to-many class="com.coginfo.kf.bean.CategoriaRuolo"/>
      </set>
    </class>

</hibernate-mapping>

<hibernate-mapping default-lazy="false">

    <class name="com.coginfo.kf.bean.CategoriaRuolo" table="KFT_CATEGORIA_RUOLO" select-before-update="true">
       <composite-id name="id" class="com.coginfo.kf.bean.CategoriaRuoloPk">
         <key-many-to-one name="codintctg"  column="codintctg" class="com.coginfo.kf.bean.Categoria"/>
         <key-many-to-one name="codintrul"  column="codintrul" class="com.coginfo.kf.bean.Ruolo"/>
       </composite-id>
    </class>
</hibernate-mapping>


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

categorieDisponibili = service.getSession().createFilter(report.getCategorie(),"where this.categoriaRuoli.id.codintrul in (" + idlistaRuoli + ")").list();

Full stack trace of any exception that occurs:
org.hibernate.QueryException: could not resolve property: codintrul of: com.coginfo.kf.bean.CategoriaRuolo [where this.categoriaRuoli.id.codintrul in (141)]

Name and version of the database you are using: Oracle 9


Could anyone help me?? It's very important! thanks


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 Post subject:
PostPosted: Fri Dec 16, 2005 9:09 am 
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Joined: Wed Sep 22, 2004 8:27 am
Posts: 89
about previous post, in hibernate conf :
<property name="hibernate.query.factory_class">ASTQueryTranslatorFactory</property>


now with this:
<property name="hibernate.query.factory_class">org.hibernate.hql.classic.ClassicQueryTranslatorFactory</property>

org.hibernate.QueryException: illegal syntax near collection: id [where this.categoriaRuoli.id.codintrul in (141)

so, i don't understand whitch hibernate.query.factory_class must use...

thanks[/b]


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 Post subject:
PostPosted: Fri Dec 16, 2005 11:11 am 
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Regular

Joined: Wed Sep 22, 2004 8:27 am
Posts: 89
anyone help me???

thanks again


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