Re: ORM Question

I think you misunderstand: I am not actually tracking llama herds. :-)

This is just example code; I haven't actually tested it in the way that you suggest.  The problem isn't with this code in particular, it's with this concept in general.  This is a question I have had again and again in the past few years - this code is just a way to articulate it.

So, in short - no, the code won't work as now written, and it's obvious why: annotate doesn't assign instances, it assigns results.

I don't think I need an annotation expert per se; it may well be that annotate is the wrong way to do this.

My SQL isn't particularly strong, but if I were writing this in SQL, I suspect that GROUP_BY is necessary or at least plausible.  I know that Django doesn't directly expose GROUP_BY, but it looks likes QuerySet.values() uses it; so I wonder if that fits into the picture somehow?

On Wed, Apr 2, 2014 at 12:35 PM, Bill Freeman <ke1g.nh@gmail.com> wrote:

My point was that annotate might work after the other problems were fixed.  Have you re-tested since?  If your annotate still doesn't work then perhaps someone with more annotate experience than I will comment.

On Wed, Apr 2, 2014 at 12:04 PM, Justin Holmes <justin@justinholmes.com> wrote:

No no, the annotate doesn't work - that's the whole reason for my question.  See, the annotate doesn't associate instances with the annotated attribute, it associated results.

(BTW, good looking out - I have once again fixed the SO question :-))

On Wed, Apr 2, 2014 at 10:17 AM, Bill Freeman <ke1g.nh@gmail.com> wrote:

1.  You now have a foreign key to Herd, but you don't have a model by that name.  It's called LlamaHerd.

2.  I was misinterpreting what you wanted (couldn't get past the lack of a relationship while reading0.  Once the relationship is fixed, your original annotate might work.

On Tue, Apr 1, 2014 at 6:20 PM, Justin Holmes <justin@justinholmes.com> wrote:

You say "use the reverse relation manager in LlamaHerd to build up your annotation," but that answer the question.

How can I sort LlamaHerd objects by the DOB of their largest llama?

On Tue, Apr 1, 2014 at 6:13 PM, Justin Holmes <justin@justinholmes.com> wrote:

Yes, absolutely - my bad.  I have updated to SO question.  I meant the models to look like:

  
class LlamaHerd(models.Model):      shepherd = ForeignKey(User)    class Llama(models.Model):      size = FloatField()      dob =
 FloatField      herd = ForeignKey(Herd, related_name="llamas")

On Tue, Apr 1, 2014 at 6:09 PM, Bill Freeman <ke1g.nh@gmail.com> wrote:

Aren't you missing a ForeignKey relationship to tell which LlamaHerd a Llama belongs to?

Then you would use the reverse relation manager in LlamaHerd to build up your annotation.

On Tue, Apr 1, 2014 at 5:52 PM, Justin Holmes <justin@justinholmes.com> wrote:

I have come across this problem in several Django projects, but only just now worked it out in my head enough to make it a cognizable question.

I have also posted it on StackOverflow, here: http://stackoverflow.com/questions/22797497/using-the-django-orm-to-order-by-a-value-in-a-distinct-related-item

Consider this example:

class LlamaHerd(models.Model):      shepherd = ForeignKey(User)    class Llama(models.Model):      size = FloatField()      dob = FloatField

How can I now make a query to get LlamaHerd objects, sorted by the date of birth of their largest llama?

It almost feels like I might be able to use annotate:

LlamaHerd.annotate(largest_llama=Max('llama__size')).order_by('largest_llama__dob')

...but the annotate here creates fields with float values, not with the model instances.

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