I'm fairly new to Python and I am using a video tutorial on Lynda to help me build the frameworks for a Social WebApp. I'm trying to run the server from the cmd using python manage.py runserver from the cmd, however, I keep running into this error message.
我对Python相当陌生，我正在使用Lynda的视频教程来帮助我构建一个社交网络应用的框架。我尝试使用python管理从cmd运行服务器。然而，来自cmd的py runserver，我一直在运行这个错误消息。
CMD PROMPT ERROR
CMD提示错误
Traceback (most recent call last):
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\utils\autoreload.py", line 225, in wrapper
fn(*args, **kwargs)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\core\management\commands\runserver.py", line 121, in inner_run
self.check(display_num_errors=True)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\core\management\base.py", line 364, in check
include_deployment_checks=include_deployment_checks,
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\core\management\base.py", line 351, in _run_checks
return checks.run_checks(**kwargs)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\core\checks\registry.py", line 73, in run_checks
new_errors = check(app_configs=app_configs)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\core\checks\urls.py", line 40, in check_url_namespaces_unique
all_namespaces = _load_all_namespaces(resolver)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\core\checks\urls.py", line 57, in _load_all_namespaces
url_patterns = getattr(resolver, 'url_patterns', [])
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\utils\functional.py", line 36, in __get__
res = instance.__dict__[self.name] = self.func(instance)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\urls\resolvers.py", line 536, in url_patterns
patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\utils\functional.py", line 36, in __get__
res = instance.__dict__[self.name] = self.func(instance)
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\urls\resolvers.py", line 529, in urlconf_module
return import_module(self.urlconf_name)
File "C:\Users\Kelechi\AppData\Local\Programs\Python\Python35\lib\importlib\__init__.py", line 126, in import_module
return _bootstrap._gcd_import(name[level:], package, level)
File "<frozen importlib._bootstrap>", line 986, in _gcd_import
File "<frozen importlib._bootstrap>", line 969, in _find_and_load
File "<frozen importlib._bootstrap>", line 958, in _find_and_load_unlocked
File "<frozen importlib._bootstrap>", line 673, in _load_unlocked
File "<frozen importlib._bootstrap_external>", line 665, in exec_module
File "<frozen importlib._bootstrap>", line 222, in _call_with_frames_removed
File "C:\Windows\SysWOW64\bookmarks\bookmarks\urls.py", line 20, in <module>
url(r'^admin/', include(admin.site.urls)),
File "C:\Users\Kelechi\AppData\Roaming\Python\Python35\site-packages\django\urls\conf.py", line 27, in include
'provide the namespace argument to include() instead.' % len(arg)
django.core.exceptions.ImproperlyConfigured: Passing a 3-tuple to include() is not supported. Pass a 2-tuple containing the list of patterns and app_name, and provide the namespace argument to include() instead.
My urls.py looks like this:
我的url。py是这样的:
from django.contrib import admin
urlpatterns = [
url(r'^admin/', include(admin.site.urls)),
...
]
1 个解决方案
#1
5
In Django 2.0 you can no longer use include(admin.site.urls) (release notes). Just use admin.site.urls instead.
在Django 2.0中，您不能再使用include(admin.site.url)(发布注释)。只使用admin.site。url。
from django.contrib import admin
urlpatterns = [
url(r'^admin/', admin.site.urls),
...
]
#1
5
In Django 2.0 you can no longer use include(admin.site.urls) (release notes). Just use admin.site.urls instead.
在Django 2.0中，您不能再使用include(admin.site.url)(发布注释)。只使用admin.site。url。
from django.contrib import admin
urlpatterns = [
url(r'^admin/', admin.site.urls),
...
]