传递参数以反向django查看

Question

我有一个视图create_rating,在我提交表单后,我希望它在视图上处理rating_upload,然后我想重定向回到create_rating视图.似乎无法使用它,我的最新代码如下.我想当我点击它应发送到submit的create-rating页面时,我可以将其作为参数发回.文档也显示了这一点.我尝试了几件事最新的错误就是我所展示的......video_idrating_uploadcreate_rating

网址:

urlpatterns = [
    url(r'^upload', UploadVideo.as_view(), name='upload'),
    url(r'^(?P<pk>[0-9]+)/$', VideoView.as_view(), name='videoview'),
    url(r'^(?P<video_id>\d+)/create_rating', create_rating, name='create_rating'),
    url(r'^(?P<video_id>\d+)/rating_upload', rating_upload, name='rating_upload'),
    url(r'^(?P<video_id>\d+)/rating_uploaded', rating_upload, name='rating_upload')
]

观点:

def create_rating(request, video_id):
    vid = get_object_or_404(Video, pk=video_id)
    past_ratings = vid.rating.order_by('date_created')[:5]
    template = loader.get_template('create_rating.html')
    context = {
        'vid': vid, 'past_ratings': past_ratings
    }
    return HttpResponse(template.render(context, request))


def rating_upload(request, video_id):
    template = loader.get_template('rating_upload.html')
    rated_video = Video.objects.get(pk=video_id)
    context = {
        'rated_video': rated_video
    }
    return HttpResponseRedirect(reverse('create_rating', video_id))

模板,create_rating.html:

<p>{{ vid.title }}</p>

<form action="{% url 'rating_upload' vid.pk %}"  method="post">

{% csrf_token %}
<input type="text" name="rate_comment">
<input type="submit" value="Rate Video">

最新错误:

Request Method: POST
Request URL: http://127.0.0.1:8000/video/32/rating_uploaded

Django Version: 1.10.5
Python Version: 2.7.10
Installed Applications:
['django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'video']
Installed Middleware:
['django.middleware.security.SecurityMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware']



Traceback:

File "/Users/RyanHelling/virtualenvs/env1/lib/python2.7/site-packages/django/core/handlers/exception.py" in inner
  39.             response = get_response(request)

File "/Users/RyanHelling/virtualenvs/env1/lib/python2.7/site-packages/django/core/handlers/base.py" in _get_response
  187.                 response = self.process_exception_by_middleware(e, request)

File "/Users/RyanHelling/virtualenvs/env1/lib/python2.7/site-packages/django/core/handlers/base.py" in _get_response
  185.                 response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "/Users/RyanHelling/PycharmProjects/flash2/video/views.py" in rating_upload
  63.     return HttpResponseRedirect(reverse('create_rating', video_id))

Exception Type: TypeError at /video/32/rating_uploaded
Exception Value: an integer is required

Answer 1

尝试

return HttpResponseRedirect(reverse('create_rating', args=(video_id,)))

代替

return HttpResponseRedirect(reverse('create_rating', video_id))

文档建议将你的args作为元组传递.