I agree with Pablo. Don't start designing queries until you've designed your
database relations, because the queries are meaningless if you change
relationships.
Sit down with a piece of paper and list or draw bubbles or whatever so that
you can establish what your models are, what the fields are in each model,
and how the models are related to one another. At that point, code them up.
Now you're ready to find the most efficient way to extract data from the
models.
If you've done this step, then in order for someone to help you design an
efficient query they would need to know how your models are related.
-Tim
On Wed, Feb 24, 2010 at 9:42 AM, Pablo Ilardi <pila...@gmail.com> wrote:
> I'd say that the effort of doing it depends on the db structure, if
> both models share tables or are under the same table hierarchy then,
> you can create a single query based on the common parent (though
> you'll only retrieve objects of the parent model and you'll have to
> determine somehow later what is the real object type to retrieve the
> missing object attributes later). If there's no common table the only
> simple sorting possibility, I'd say, is to sort then in memory.
>
> - Pablo
>
> On Feb 24, 11:40 am, Jirka Vejrazka <jirka.vejra...@gmail.com> wrote:
> > Hi,
> >
> > > i have a simple question here, i've been trying to found the answer
> > > but somehow i got confused if this is wheter possible or not:
> >
> > it's always possible, just a matter of time and effort (and cost) ;-)
> >
> > > I have two django models, movies and books , i want to get all the
> > > objects from both models and order them by date, getting result_set
> > > that
> > > might look like this:
> >
> > > movie1
> > > movie2
> > > book1
> > > movie3
> > > book2
> >
> > It depends a bit on the context (i.e. do you really want all objects
> > or just the most recent ones?). I'll only provide a few hints.
> >
> > - convert whatever you need to combine into lists. You may want all
> > or just latest 20 objects, i.e
> >
> > >>> movies = Movie.objects.all()
> > >>> books = Book.objects.all()
> >
> > or
> >
> > >>> movies = Movie.objects.all().order_by('-id')[:20]
> > >>> books = Book.objects.all().order_by('-id')[:20]
> >
> > then
> >
> > >>> my_objects = list(movies).extend(list(books))
> >
> > then
> >
> > >>> my_objects.sort(key=...) # seehttp://
> wiki.python.org/moin/HowTo/Sorting
> >
> > Hope this helps
> >
> > Jirka
>
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