On Aug 7, 7:47 pm, rmschne <rmsc...@gmail.com> wrote:
> I have a model like that I would like query down more than one level
> (using my nomenclature).
>
> class Library(models.Model)
> branch=models.CharField(max_length=100)
>
> class Book(models.Model)
> libraryid=models.ForeignKey(Library,related_name="books")
>
> class Chapter(models.Model)
> bookid=models.ForeignKey(Book,related_name="chapters")
>
> I can get a library object:
>
> l=Library.objects.filter(branch="Covent Garden")
>
> and then get a query set of all the books in Covent Garden
>
> b=l.books.all()
>
> How do I get a query set of all the chapters (in all the books) in the
> Covent Garden library?
>
> then then if I added
>
> class Paragraph(models.Model)
> chapterid=models.ForeignKey(Chapter,related_name="paragraphs")
>
> How would I get a query set of all the paragraphs (in all the chapters
> in all the books) in the Covent Garden library?
Use the double-underscore syntax to traverse relationships. For all
chapters in books in Covent Garden:
Chapter.objects.filter(bookid__libraryid__branch='Covent Garden')
and all paragraphs in all chapters in all books in Covent Garden:
Paragraph.objects.filter(chapterid__bookid__libraryid__branch='Covent
Garden')
and so on.
Two things, though: firstly, your example query `b=l.books.all()` will
*not* work. `l` is a queryset, and doesn't have a `books` attribute.
If you're starting with an existing queryset, you can use `in`:
b = Books.objects.get(libraryid__in=l)
but in this case there's presumably only one Covent Garden branch, so
you should have just used `get` to get a single instance, rather than
a queryset, in the first place:
l = Library.objects.get(branch='Covent Garden')
and then your `l.books.all()` would have worked.
Secondly, please drop those `id` suffixes. libraryid and bookid are
not ID fields, they are fields representing whole objects in other
models. Django automatically names the underlying database field
foo_id, so you will have libraryid_id and bookid_id behind the scenes.
--
DR.
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