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python – 使用include(‘admin.site.urls’)时出错:不支持将3元组传递给include()

2019-08-23 13:58:59  阅读:247  来源: 互联网

标签:python django django-2-0 valueerror


我是Python的新手,我在Lynda上使用视频教程来帮助我构建Social WebApp的框架.我正在尝试使用cmd中的python manage.py runserver从cmd运行服务器,但是,我一直遇到此错误消息.

CMD PROMPT ERROR

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.

我的urls.py看起来像这样:

from django.contrib import admin

urlpatterns = [
    url(r'^admin/', include(admin.site.urls)),
    ...
]

解决方法:

在Django 2.0中,您不能再使用include(admin.site.urls)(release notes).只需使用admin.site.urls即可.

from django.contrib import admin

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    ...
]

标签:python,django,django-2-0,valueerror
来源: https://codeday.me/bug/20190823/1698009.html

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