Question

我正在与django合作，我试图让一个简单的应用程序正常运行。

该应用应该提供一个简单的“Hello world”视图，但my_app/urls.py无法从my_app/views.py导入方法。

我的应用名为 SocialUrl

这是my-project/urls.py。

from django.conf.urls import patterns, include, url
# from SocialUrl.views import test

from django.contrib import admin
admin.autodiscover()

    urlpatterns = patterns('',
    url(r'^admin/', include(admin.site.urls)),
    url(r'^/su/', include('SocialUrl.urls')),
    # url(r'^$', test),
)

以下是my-app/urls.py

from django.conf.urls import patterns, url
from . import views

urlpatterns = patterns('',
    # ex: /polls/
    url(r'^test$', views.test()),
)

以下是my-app/views.py

from django.shortcuts import render
from django.template import RequestContext

def test(request):
    return render(request, 'test.html', RequestContext(request, {}))

注意： - my-project/urls.py中的测试行正常运行。

我的文件夹结构如下所示：

my-project
|- my-project
|- my-app

我已在my-project/settings.py

中添加了此内容

TEMPLATE_DIRS = (
    os.path.join(os.path.dirname(__file__), 'templates').replace('\\', '/'),
    os.path.join(BASE_DIR, 'SocialUrl', 'templates').replace('\\', '/'),
)

我收到以下错误：

    Environment:


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

    Django Version: 1.6.2
    Python Version: 2.7.6
    Installed Applications:
    ('django.contrib.admin',
     'django.contrib.auth',
     'django.contrib.contenttypes',
     'django.contrib.sessions',
     'django.contrib.messages',
     'django.contrib.staticfiles',
     'SocialUrl')
    Installed Middleware:
    ('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 "C:\Users\Sverker\.virtualenvs\coomba\lib\site-packages\django\core\handlers\base.py" in get_response
      101.                 resolver_match = resolver.resolve(request.path_info)
    File "C:\Users\Sverker\.virtualenvs\coomba\lib\site-packages\django\core\urlresolvers.py" in resolve
      318.             for pattern in self.url_patterns:
    File "C:\Users\Sverker\.virtualenvs\coomba\lib\site-packages\django\core\urlresolvers.py" in url_patterns
      346.         patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module)
    File "C:\Users\Sverker\.virtualenvs\coomba\lib\site-packages\django\core\urlresolvers.py" in urlconf_module
      341.             self._urlconf_module = import_module(self.urlconf_name)
    File "C:\Users\Sverker\.virtualenvs\coomba\lib\site-packages\django\utils\importlib.py" in import_module
      40.         __import__(name)
    File "C:\Users\Sverker\Dropbox\Coomba\SbrgCoomba\SbrgCoomba\urls.py" in <module>
      13.     url(r'^/su/', include('SocialUrl.urls')),
    File "C:\Users\Sverker\.virtualenvs\coomba\lib\site-packages\django\conf\urls\__init__.py" in include
      26.         urlconf_module = import_module(urlconf_module)
    File "C:\Users\Sverker\.virtualenvs\coomba\lib\site-packages\django\utils\importlib.py" in import_module
      40.         __import__(name)
    File "C:\Users\Sverker\Dropbox\Coomba\SbrgCoomba\SocialUrl\urls.py" in <module>
      7.     url(r'^$', test),

    Exception Type: NameError at /
    Exception Value: name 'test' is not defined

我在from . import views尝试了my-app/urls.py行的各种组合例如。 from views import test

编辑：

仍然失效my-app/urls.py

from django.conf.urls import patterns, url
from views import test

urlpatterns = patterns('',
    # ex: /polls/
    url(r'^$', test),
)

Answer 1

问题是你的应用urls.py确实有这一行：

url(r'^$', test)

错误表明test未定义。只需定义（导入）它：

from views import test

Answer 2

我认为这可能会导致错误：

url(r'^/su/', include('SocialUrl.urls')),

删除前导斜杠，您将只有：

url(r'^su/', include('SocialUrl.urls')),

请告诉我这是否适合您

Answer 3

我不确定这是否是问题 - 但通常我会按照此模式查看我的应用级网址文件。

主要项目urls.py文件

from django.conf.urls import patterns, include, url

urlpatterns = patterns('apps.main.views',
    url(r'', include('apps.main.urls')),
)

应用级别urls.py文件

from django.conf.urls import patterns, url

urlpatterns = patterns('',
    # ex: /polls/
    url(r'^test$', 'views.tests'),
)

请注意，您应该使用urlpatterns +=或urlpatterns =，具体取决于之前是否已定义。

从views.py导入无法从应用程序

3 个答案: