Patrick Burns Patrick Burns - 4 months ago 25
Python Question

Django - Dictionary update sequence element #0 has length 1; 2 is required

After i created new URLs, i got this error

ValueError at /blog/ dictionary update sequence element #0 has length 1; 2 is required

Traceback

Request Method: GET
Request URL: http://127.0.0.1:8000/blog/

Django Version: 1.7.2
Python Version: 3.4.1
Installed Applications:
('django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
'modeltranslation',
'blog')
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 "C:\Python34\lib\site-packages\django\core\handlers\base.py" in get_response
98. resolver_match = resolver.resolve(request.path_info)
File "C:\Python34\lib\site-packages\django\core\urlresolvers.py" in resolve
345. sub_match = pattern.resolve(new_path)
File "C:\Python34\lib\site-packages\django\core\urlresolvers.py" in resolve
345. sub_match = pattern.resolve(new_path)
File "C:\Python34\lib\site-packages\django\core\urlresolvers.py" in resolve
222. kwargs.update(self.default_args)

Exception Type: ValueError at /blog/
Exception Value: dictionary update sequence element #0 has length 1; 2 is required


I use this urls

# /project/urls.py
from blog.urls import urlpatterns as blog_urls

urlpatterns = patterns('',
url(r'^admin/', include(admin.site.urls)),
url(r'^blog/', include(blog_urls, 'blog')),
)

# /blog/urls.py
urlpatterns = patterns('blog.views',
url(r'^$', 'index', 'index'),
url(r'^category/(?P<category_id>\d+)/$', 'category', 'category'),
)

# /blog/views.py
def index(request):
render(request, 'blog/index.html')

def category(request, category_id):
render(request, 'blog/index.html')


Why i got this error? Never saw it before and i don't know why i getting this error...

Answer

The problem is this line:

url(r'^category/(?P<category_id>\d+)/$', 'category', 'category')

The third positional parameter is the extra context dict. If you want to pass the name instead, you need to use a keyword arg:

url(r'^category/(?P<category_id>\d+)/$', 'category', name='category')