Question

我有选择字段。我想要这样的views.py; 我该怎么办？

if gametype == 1: 
    template='xxx.html'

if gametype == 2:
    template='xxx1.html'

views.py

def game_detail(request,tournamentslug,slug,gameslug):
    game=get_object_or_404(Model,tournament__slug=tournamentslug,slug=slug,game__slug=gameslug)

    context={
        'game':game,
    }

    return render(request,'esports/lolgame.html',context)

models.py

class Model(models.Model):
    type_tvt = 1
    type_pvp = 2
    type_royale=3
    types = (
        (type_tvt, 'T'),
        (type_pvp, 'P'),
        (type_royale,'R'),
    )
    gametype=models.SmallIntegerField(choices=types)

Answer 1

假设您有一个基于类的详细信息视图（出于完整性考虑，该模型被命名为Game，而不是示例中的Model），

class GameView(generic.DetailView):
    model = Game

    def get_template_names(self):
        if self.object.gametype == Game.type_tvt:
            return ['xxx.html']
        elif self.object.gametype == Game.type_pvp:
            return ['xxx2.html']
        elif self.object.gametype == Game.type_royale:
            return ['xxx3.html']
        raise ValueError('invalid game type')

可以解决问题–或简化使用字典的方法，

class GameView(generic.DetailView):
    model = Game
    template_names = {
         Game.type_tvt: 'xxx.html',
         Game.type_pvp: 'xxx2.html',
         Game.type_royale: 'xxx3.html',
    }

    def get_template_names(self):
        return [self.template_names[self.object.gametype]]  # may raise KeyError

编辑：对于基于功能的视图，如在已编辑的问题中一样，

template_names = {
    Model.type_tvt: "xxx.html",
    Model.type_pvp: "xxx2.html",
    Model.type_royale: "xxx3.html",
}


def game_detail(request, tournamentslug, slug, gameslug):
    game = get_object_or_404(
        Model,
        tournament__slug=tournamentslug,
        slug=slug,
        game__slug=gameslug,
    )

    context = {"game": game}
    template_name = template_names[game.gametype]
    return render(request, template_name, context)

Django查看if语句

1 个答案: