python - Django select_lated 通过单个查询连接模型属性

Question

我正在尝试通过单个查询找到最佳解决方案来检索连接模型上的属性。

我有以下模型关系:

class Player(models.Model):
    first_name = models.CharField(max_length=30)
    last_name = models.CharField(max_length=30)
    season_team = models.ForeignKey(SeasonTeam, related_name='players', blank=True, null=True)
    leagues = models.ManyToManyField(League, related_name='players', through='PlayerLeague')


class League(models.Model):
    name = models.CharField()


class PlayerLeague(models.Model):
    player = models.ForeignKey(Player)
    league = models.ForeignKey(League)
    stamina = models.PositiveSmallIntegerField(_('Stamina'), default=100)


class Team(models.Model):
    name = models.CharField(max_length=30, validators=[
        RegexValidator(regex='^[a-zA-Z0-9\s]+$', message=_('name must contain only chars and numbers'), ), ])
    players = models.ManyToManyField(Player, blank=True, related_name='teams')
    league = models.ForeignKey(League, related_name='teams')

这是我希望在 get_context_data 中加载的所有玩家预加载耐力属性的 View

class FormationUpdateView(AjaxResponseMixin, FormationRequirementsMixin, UpdateView):
    model = Formation
    template_desktop = 'hockey/formation/form.html'
    template_mobile = 'hockey/formation/mobile/form.html'
    form_class = FormationForm

    def get_initial(self):
        self.initial = {'active': True}
        return self.initial.copy()

    def get_context_data(self, **kwargs):
        context = super(FormationUpdateView, self).get_context_data(**kwargs)
        team = Team.objects.select_related('league').get(formations=self.kwargs['pk'])
        context['players'] = Player.objects.filter(teams=team).all()
        context['league'] = team.league
        return context

    def get_queryset(self):
        return Formation.objects.select_related()

    def get_template_names(self):
        if 'Mobile' in self.request.META['HTTP_USER_AGENT']:
            return self.template_mobile
        return self.template_desktop

我的第一个解决方案是向 Player 类添加 stamina 方法:

def stamina(self, league):
    p_l = self.leagues.through.objects.filter(league=league)[0]
    return p_l.stamina

我不喜欢这个解决方案，因为它将查询每个玩家的耐力值。

感谢您的帮助

Answer 1

你的模型有点令人费解，但类似的东西应该可以最大限度地减少请求数量

from django.db.models import Q, F, Prefetch

... 

def get_context_data(self, **kwargs)
     context = super(FormationUpdateView, self).get_context_data(**kwargs)
     team = (Team.objects
     .select_related('league')
     .prefetch_related(
         Prefetch(
             'players',
             queryset=(Player.objects.filter(leagues=F('leagues'))
                       .prefetch_related('playerleagues'))
         )
     ).get(formations=self.kwargs['pk']))
     context['players'] = team.players.all()
     context['league'] = team.league
     return context

无需再次访问数据库即可获取玩家的体力。

players[0].playerleagues.all()[0].stamina