Thanks Kevin, you got me on the right track. I was able to implement the
following solution:
records = models.Facility.objects.filter(...)
for x in records:
schedule = x.schedule_set.all()
But I could not find a way to do it all at once like
records = models.Facility.objects.filter(...).schedule_set.all()
If there is not a "clean" method to do this, I'm inclined to use the above
solution since I have only ~100 records in the first query. I don't want to
"drop into SQL" unnecessarily. On this query, I probably only need the
.count() method on schedule_set() if that even makes a difference.
-----Original Message-----
From: django-users@googlegroups.com [mailto:django-users@googlegroups.com] On
Behalf Of Kevin
Sent: Tuesday, October 11, 2011 8:43 AM
To: Django users
Subject: Re: Need advice on ForeignKey query problem.
What your looking for is here in the documentation:
https://docs.djangoproject.com/en/1.3/topics/db/queries/#related-objects
records = models.Facility.objects.get(pk=62).schedule_set.all()
On Oct 10, 8:19 pm, "Sells, Fred" <fred.se...@adventistcare.org>
wrote:
> I've got these two tables defined where a Facility can have multiple
> schedules but a schedule can have only one facility.
>
> class Facility(models.Model):
> id = models.CharField(max_length=2, primary_key=True)
> name = models.CharField(max_length=30)
> ....
>
> class Schedule(models.Model):
> facility = models.ForeignKey(Facility)
> ...
>
> I would like a query that returns one record for each Facility and that
> record would contain at least the id (if not more) of ALL linked
> schedules. So far I've only been able to return a queryset that
> contains a record for each schedule at each facility. My last attempt
> looked like this...
>
> records = models.Facility.objects.select_related().filter(id='62')
>
> is there a pythonic way to do this? I only need reasonable efficiency,
> there's not that much data and the query is probably only run 200 times
> in an 8 hour shift.
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