I should pay better attention, try:
Tours.objects.all().order_by('country')
-richard
On 9/24/07, Oleg Korsak <[EMAIL PROTECTED] > wrote:
>
> Error in formatting: отсутствует запись о FROM-выражении для таблицы
> "tours_country" LINE 1: ..."tours_tour"."user_id" FROM "tours_tour"
> ORDER BY "tours_cou... ^
>
> It says that there is no tours_country in FROM statement...
>
> Richard Dahl пишет:
> > Am I missing something or wouldn't
> > Tours.objects.all().order_by('tours_country.name')
> > work?
> > -richard
> >
> >
> > On 9/24/07, Oleg Korsak <[EMAIL PROTECTED] > wrote:
> >> Hello! I have such models:
> >>
> >> class Country(models.Model):
> >> name = models.CharField(_('name'), maxlength=32)
> >>
> >> def __unicode__(self):
> >> return self.name
> >>
> >> class Meta:
> >> verbose_name = _('country')
> >> verbose_name_plural = _('countries')
> >> ordering = ('name',)
> >>
> >> class Admin:
> >> pass
> >>
> >>
> >> class Tour(models.Model):
> >> id = models.AutoField(primary_key=True)
> >> country = models.ForeignKey(Country)
> >> name = models.CharField(_('name'),
> >> maxlength=64,blank=False,null=False)
> >> user = models.ForeignKey
> (User,blank=True,null=True,default=None)
> >>
> >> def __unicode__(self):
> >> return self.name
> >>
> >> class Meta:
> >> verbose_name = _('tour')
> >> verbose_name_plural = _('tours')
> >> ordering = ('name',)
> >> order_with_respect_to = 'country'
> >>
> >> class Admin:
> >> pass
> >>
> >> I want to get such result:
> >> SELECT c.name,t.name FROM tours_tour t, tours_country c WHERE
> >> t.country_id=c.id ORDER BY c.name;
> >>
> >> For now I've wrote such dirty workaround:
> >>
> >> Tour.objects.extra(where=['tours_country.id=tours_tour.country_id'],
> >> tables=['tours_country']).order_by('tours_country.name')
> >>
> >> BUT! I don't want to use table names and so on! how can I use django
> >> model api to do this instead of writing such workarounds? I need to get
>
> >> all tours sorted by country name.
> >>
> >> Thanks.
> >>
> >>
> >>
> >
> > > >
> >
>
>
>
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