Re: [M-R] order by on custom SELECT column
On 5/1/06, Cheng Zhang <[EMAIL PROTECTED]> wrote:
> I use a custom SELECT column as:
> select = {
> 'choices': 'SELECT COUNT(*) FROM polls_choice WHERE
> poll_id=polls_poll.id',
> }
> p = Poll.objects.extra(select=select).order_by('choices')
>
> By trying this, I found out a bug in the ORM layer which I reported
> and submitted a patch as:
> http://code.djangoproject.com/ticket/1730
>
> Custom SELECT column is one way, I'd like to know is there any other
> ways to get the same thing, esp. only using db api since I try to
> avoid doing custom SQL as much as possible for portability purpose.
Hi Cheng,
There's no cleaner way to get the same thing, but I think that would
be a fine addition. Maybe something like
Poll.objects.populate_related_aggregates('choice', 'count').get(id=3).
Syntax would be the main thing to figure out here.
Adrian
--
Adrian Holovaty
holovaty.com | djangoproject.com
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[M-R] order by on custom SELECT column
On the models of tutorial 1, I want to get a list of Poll, order by
how many choices it has.
class Poll(models.Model):
question = models.CharField(maxlength=200)
pub_date = models.DateTimeField('date published')
class Choice(models.Model):
poll = models.ForeignKey(Poll)
choice = models.CharField(maxlength=200)
votes = models.IntegerField()
I use a custom SELECT column as:
select = {
'choices': 'SELECT COUNT(*) FROM polls_choice WHERE
poll_id=polls_poll.id',
}
p = Poll.objects.extra(select=select).order_by('choices')
By trying this, I found out a bug in the ORM layer which I reported
and submitted a patch as:
http://code.djangoproject.com/ticket/1730
Custom SELECT column is one way, I'd like to know is there any other
ways to get the same thing, esp. only using db api since I try to
avoid doing custom SQL as much as possible for portability purpose.
BR,
- Cheng
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