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...@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 <jus...@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.
>>
>>
>> --
>> Justin Holmes
>> Chief Chocobo Breeder, slashRoot
>>
>> slashRoot: Coffee House and Tech Dojo
>> New Paltz, NY 12561
>> 845.633.8330
>>
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--
Justin Holmes
Chief Chocobo Breeder, slashRoot
slashRoot: Coffee House and Tech Dojo
New Paltz, NY 12561
845.633.8330
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