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 Post subject: Manipulate SQLQuery results
PostPosted: Wed Aug 01, 2012 9:06 am 
Newbie

Joined: Wed Aug 01, 2012 5:51 am
Posts: 8
Hi !

According to my browsing, getting the results from a SQLQuery should look like this :
Code:
         session = HibernateUtil.currentSession();

         String query = "SELECT num_loc FROM Circulation";
         
         List<Object[]> rows = (List<Object[]>)session.createSQLQuery(query)
         .addScalar("num_loc", Hibernate.STRING)
         .list();
         
         System.out.println(rows.size());
         
         for (Object[] row : rows){
            System.out.print(row[0]);
         }

But I get an error when it comes to the "row[0]". Do you where I'm doing wrong ? I'm using Hibernate 3 by the way !

Thanks in advance !


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 Post subject: Re: Manipulate SQLQuery results
PostPosted: Mon Aug 06, 2012 4:32 am 
Newbie

Joined: Wed Aug 01, 2012 5:51 am
Posts: 8
For those who may have the same question, this works fine in my program :
Code:
      try {
         session = HibernateUtil.currentSession();

         String num = "XXXX";

         String sql_query = "SELECT C.Id, C.num_loc, A.Lieux FROM Circulation C, Appro_gasoil A WHERE C.num_loc = num";
         
         @SuppressWarnings("unchecked")
         Query query = session.createSQLQuery(sql_query)
         .addScalar("Id", Hibernate.INTEGER)
         .addScalar("num_loc", Hibernate.STRING)
         .addScalar("Lieux", Hibernate.STRING);
         
         List result = (List)query.list();
         
         for(int i=0; i<result.size(); i++) {
            Object[] ligne = (Object[])result.get(i);
            System.out.println(ligne[2]);
         }
      
      } catch (HibernateException e) {
         System.out.println("Erreur dans la requĂȘte concernant les approvisionnements en gasoil");
         e.printStackTrace();
      }


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