Django将自定义表单参数传递给ModelFormset

Question

我的问题类似于 Django将自定义表单参数传递给Formset

我有这些课程

class Game(models.Model):
    home_team = models.ForeignKey(Team, related_name='home_team')
    away_team = models.ForeignKey(Team, related_name='away_team')
    round = models.ForeignKey(Round)

TEAM_CHOICES = ((1, '1'), (2, 'X'), (3, '2'),)

class Odds(models.Model):
    game = models.ForeignKey(Game, unique=False)
    team = models.IntegerField(choices = TEAM_CHOICES)
    odds = models.FloatField()
    class Meta:
        verbose_name_plural = "Odds"
        unique_together = (
            ("game", "team"),
        )

class Vote(models.Model):
    user = models.ForeignKey(User, unique=False)
    game = models.ForeignKey(Game)
    score = models.ForeignKey(Odds)
    class Meta:
        unique_together = (
            ("game", "user"),)

我已经定义了自己的modelformset_factory：

def mymodelformset_factory(ins):
    class VoteForm(forms.ModelForm):
        score = forms.ModelChoiceField(queryset=Odds.objects.filter(game=ins), widget=forms.RadioSelect(), empty_label=None)
        def __init__(self, *args, **kwargs):
            super(VoteForm, self).__init__(*args, **kwargs)
        class Meta:
            model = Vote
            exclude = ['user']
    return VoteForm 

我这样使用它：

        VoteFormSet = modelformset_factory(Vote, form=mymodelformset_factory(v), extra=0)
        formset = VoteFormSet(request.POST, queryset=Vote.objects.filter(game__round=round, user=user))

显示表单：

指定回合中的游戏下拉框，并且应该为赔率显示3个单选按钮，但我不知道将什么作为参数传递给mymodelformset_factory ..如果v = Game.objects。得到（pk = 1）它显然只显示所有游戏的pk = 1的游戏，我需要的是v = Game.objects.get（pk =“游戏谁连接到关于”的赔率“）如果你抓住了我的漂移..

Answer 1

我认为您想对自定义工厂功能进行一些更改。它应该返回formset类，而不是表单。怎么样：

def make_vote_formset(game_obj, extra=0):
    class _VoteForm(forms.ModelForm):
        score = forms.ModelChoiceField(
            queryset=Odds.objects.filter(game=game_obj), 
            widget=forms.RadioSelect(), 
            empty_label=None)

        class Meta:
            model = Vote
            exclude = ['user',]
    return modelformset_factory(Vote, form=_VoteForm, extra=extra)

然后在您的视图代码中：

current_game = Game.objects.filter(id=current_game_id)
VoteFormSet = make_vote_formset(current_game)
formset = VoteFormSet(
             request.POST, 
             queryset=Vote.objects.filter(game__round=round, user=user))

Answer 2

另一种解决方案是继承BaseModelFormSet并覆盖 _construct_forms 方法。默认情况下，BaseFormSet _construct_form方法仅使用一个参数调用_construct_forms，i：

# django/forms/formsets.py
def _construct_forms(self):
    # instantiate all the forms and put them in self.forms
    self.forms = []
    for i in xrange(self.total_form_count()):
        self.forms.append(self._construct_form(i))

但可以有任意数量的关键字参数：

# django/forms/formsets.py
def _construct_form(self, i, **kwargs):

所以，这是我的view方法和表单，它从_construct_form接收 init 中的附加参数：

# view

def edit(request, id):
    class ActionsFormSet(BaseModelFormSet):
        department = request.user.userdata.department.pk

        def _construct_forms(self):
            self.forms = []
            for i in range(self.total_form_count()):
                self.forms.append(self._construct_form(i, dep=self.department))

    actions_formset = modelformset_factory(Action, form=ActionForm, extra=0, can_delete=True, formset=ActionsFormSet)
    ...


# form

class ActionForm(forms.ModelForm):
    def __init__(self, dep=0, *args, **kwargs):
        super(ActionForm, self).__init__(*args, **kwargs)
        self.fields['user'].choices = [(u.pk, u.first_name) for u in User.objects.filter(userdata__department=dep)]

    class Meta:
        model = Action
        exclude = ('req',)