Question

如果我正在使用我的views.py看起来像的模型表单：

def dog_image_upload(request):
    if request.user.is_authenticated():
        if request.method == 'POST':
            form = DogImageForm(request.POST, request.FILES)
            if form.is_valid():
                form.save()
        else:
            form = DogImageForm(user)
        return render_to_response("dog-image-upload.html", {'form': form}, context_instance=RequestContext(request))
    else:
        return HttpResponseRedirect('/')

在model.py中，我想这样做：

class DogImageForm(ModelForm):
    dogs = forms.ModelChoiceField(queryset=Dog.objects.filter(user=request.user))
    class Meta:
        model = ResultsUpload
        fields = ['dogs','image']

但是，我在尝试将用户发送到model.py时遇到问题对此的帮助将是非常棒的，值得一试！

Answer 1

您必须在模型中使用__init__

class DogImageForm(ModelForm):
    dogs = forms.ModelChoiceField(queryset=Dog.objects.none())
    class Meta:
        model = ResultsUpload

    def __init__(self, user, *args, **kwargs):
        super(DogImageForm, self).__init__(*args, **kwargs)
        self.fields['dogs'].queryset = Dog.objects.filter(user=user)

并在表单初始化期间

form = DogImageForm(user=request.user)

访问modelForm中的request.user

1 个答案: