On 10/25/06, Christian Joergensen <[EMAIL PROTECTED]> wrote:
> Daniel Roseman wrote:
> >>> Country.objects.all().extra(
> >>> where=['(SELECT COUNT(*) FROM appname_soccerteam WHERE
> >>> appname_soccerteam.country_id = appname_country.id) > 0']
> >>> )
> >> Great idea! I did not think about the 'extra' method. Works perfectly.
> >> Gave me a speedup factor of about 250k :)
> >>
> >> Thanks,
> >
> > Alternatively you could use the normal django syntax:
> > Country.objects.filter(soccerteam__name__isnull=False)
>
> But the thing is, I can have multiple SoccerTeam relations. Doesn't that
> rule out this solution?
You can use distinct():
Country.objects.filter(soccerteam__name__isnull=False).distinct()
Country won't have a soccerteam attribute - by default, it will have a
soccerteam_set attribute. If you add a related_name argument to the
ForeignKey in the SoccerTeam class, you can change the name of this to
whatever you want.
class SoccerTeam(models.Model):
...
country = models.ForeignKey(Country, related_name='soccerteams')
...
Country.objects.filter(soccerteams__id__isnull=False).distinct()
I'd guess the INNER JOIN the ORM for you (the check for a particular
field being null is bogus in this case - it's the join which does the
work for you) would be faster than the COUNT.
Jonathan.
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