Question

嗨，这应该是一个简单的我确定。只是失去了对基于类的观点的耐心。不确定他们如何简化事情。基于函数的视图似乎更具可读性并且更有意义。无论如何我试图将我的一个功能视图转换为一个类，并在我请求视图时收到此错误

'tuple'对象没有属性'get'

PS我正在使用django 1.8.1，我导入View的方式是通过django.views.generic.View而不是django.views.View，如文档中所述。可能是相关的。

这是urls.py

from django.conf.urls import url
from . import views


urlpatterns = [
    url(r'^upload_photo/$', views.upload_photo, name='upload_photo'),
    url(r'^upload_photo_class/$', views.UploadPhotoView.as_view()),
    ]

以下是基于类的视图

class UploadPhotoView(View):
    form = PhotoModelForm
    formset = formset_factory(PhotoExtendedModelForm, max_num=1, validate_min=True)
    template_name = 'upload_photo.html'

    @method_decorator(login_required)
    def dispatch(self, *args, **kwargs):
        return super().dispatch(*args, **kwargs)

    def get(self, request):
        form = self.form
        formset = self.formset
        return (request, self.template_name, {'form': form,
                                          'formset': formset})

    def post(self, request, *args, **kwargs):
        form = self.form(request.POST, request.FILES)
        formset = self.formset(request.POST)
        if all([form.is_valid(), formset.is_valid()]):

            photo = form.save()
            photo_instance = photo
            user_id = request.user

            for inline_form in formset:

                data = inline_form.save(commit=False)
                data.Photo = photo_instance
                data.user = user_id
                data.save()

            return render(request, 'upload_photo_done.html')

        else:
            return (request, self.template_name, {'form': form,
                                              'formset': formset})

这是基于功能的原始视图

@login_required
def upload_photo(request):
    form = PhotoModelForm
    formset = formset_factory(PhotoExtendedModelForm, max_num=1, validate_min=True)

    if request.method == 'POST':
         form = form(request.POST, request.FILES)
         formset = formset(request.POST)
         if all([form.is_valid(), formset.is_valid()]):

            photo = form.save()
            photo_instance = photo
            user_id = request.user

            for inline_form in formset:

                data = inline_form.save(commit=False)
                data.Photo = photo_instance
                data.user = user_id
                data.save()

            return render(request, 'upload_photo_done.html')

        else:
            return render(request, 'upload_photo.html', {'form': form, 'formset': formset})
    else:
        return render(request, 'upload_photo.html', {'form': form, 'formset': formset})

继承人追溯：

Request Method: GET
Request URL: http://127.0.0.1:8000/otologue/upload_photo_class/

Django Version: 1.8.1
Python Version: 3.4.4
Installed Applications:
('account',
 'photologue_custom',
 'otologue',
 'photologue',
 'sortedm2m',
 'django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.sites',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'django.contrib.sitemaps')
Installed Middleware:
('django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.auth.middleware.SessionAuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware')


Traceback:
File "../venv/lib/python3.4/site-packages/django/core/handlers/base.py" in get_response
  223.                 response = middleware_method(request, response)
File "../venv/lib/python3.4/site-packages/django/middleware/clickjacking.py" in process_response
  31.         if response.get('X-Frame-Options', None) is not None:

Exception Type: AttributeError at /otologue/upload_photo_class/
Exception Value: 'tuple' object has no attribute 'get'

Answer 1

在def get（）中，你应该像这样返回一个 HttpResponse 对象

     HttpResponseRedirect('/success/')

或使用render（）重定向到模板。您正在尝试使用渲染函数调用参数返回一个普通元组。

Answer 2

您希望继承having min(status = 'Y' then 1 else 0 end) = 1而非FormView。因此，您的第一行将是view

它将简化大量的表单处理和样板，这是基于类的视图的很多好处。像往常一样，Django documentation on this is很棒。

在您的情况下，可能就像：

一样简单

class UploadPhotoView(FormView):

请求基于类的视图Django时，'tuple'对象没有属性'get'

2 个答案: