Thanks Micheal for the response!
But your query counts all the books, even the bad ones. I only need to
count the good ones...
For example, if my books are:
1. name: LOTR, rating: 10, publisher: A ...
2. name: ASOIAF, rating: 10, publisher: A ...
3. name: Twilight, rating 1, publisher: B ...
and my publishers are:
A and B
Your query returns:
[A, num_book=2]
[B, num_book=1]
the query on my first message returns:
[A, num_book=2]
and what I need is:
[A, num_book=2]
[B, num_book=0]
where num_books means number of good books.
I'm trying to do this with a single query because i need to iterate
over the queryset in one template with something like:
{% for p in publishers %}
{{ p.name }} - {{ p.number_good_books }}
{% endfor %}
Right now I'm using a custom method in the Publisher model
( get_number_of_good_books() ).
It works but it's really slow (one query for each publisher to count
the number of good books).
Ciao
On Feb 21, 7:50 pm, Michael Elkins <michael.elk...@gmail.com> wrote:
> On Feb 21, 6:11 am, Enrico <poe...@gmail.com> wrote:
>
> > This query:
> > Publisher.objects.filter(book__rating__gt=3.0).annotate(num_books=Count('bo
> > ok'))
> > returns all the publishers with at least one good book (ranked > 3)
> > with annotated the number of good books for each publisher.
>
> > How can i modify the query to get the list of ALL publishers with
> > annotated the number of good books for each publisher?
> > In other words I want to keep in the results also the Publishers
> > without good books (num_books = 0).
>
> Hi Enrico,
>
> The Django documentation has an example of what you are trying to do:
>
> https://docs.djangoproject.com/en/1.3/topics/db/aggregation/#generati...
>
> The answer to your specific question is to do it this way:
>
> qs = Publisher.objects.annotate(num_books=Count('book'))
> for o in qs:
> print o.num_books
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