Django - 字典更新序列元素＃0的长度为1；需要2

Question

我创建了新的URL后，出现了以下错误：

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

我使用这些URL链接

# /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')

为什么我会收到这个错误？以前从未见过，也不知道为什么会出现这个错误...

- Patrick Burns

Answer 1

问题出在这行代码：

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

第三个位置参数是额外的上下文字典。如果您想传递名称，您需要使用关键字参数：

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

- Daniel Roseman

---

谢谢，添加 'name=' 就解决了问题！ - Dror

任何人都可能忘记的小细节，却会浪费很多时间来调试它们！！ - Mohammad Jafar Mashhadi

谢谢，这真的帮我省了很多时间。 - Nihal Sharma