我目前正在尝试django。我在urls.py namespace中的一个参数中使用了参数。include()当我运行服务器并尝试浏览时,出现此错误。
File "C:\Users\User\AppData\Local\Programs\Python\Python36-32\lib\site-packages\django\urls\conf.py", line 39, in include 'Specifying a namespace in include() without providing an app_name ' django.core.exceptions.ImproperlyConfigured: Specifying a namespace in include() without providing an app_name is not supported. Set the app_name attribute in the included module, or pass a 2-tuple containing the list of patterns and app_name instead.
这些是我的urls.py文件:
#project/urls.py from django.conf.urls import include, url from django.contrib import admin urlpatterns = [ url(r'^reviews/', include('reviews.urls', namespace='reviews')), url(r'^admin/', include(admin.site.urls)), ]
和
#app/urls.py from django.conf.urls import url from . import views urlpatterns = [ # ex: / url(r'^$', views.review_list, name='review_list'), # ex: /review/5/ url(r'^review/(?P<review_id>[0-9]+)/$', views.review_detail, name='review_detail'), # ex: /wine/ url(r'^wine$', views.wine_list, name='wine_list'), # ex: /wine/5/ url(r'^wine/(?P<wine_id>[0-9]+)/$', views.wine_detail, name='wine_detail'), ]
我如何通过app_name错误消息中所述的?
你所做的不是传递参数以包含的可接受方法。你可以这样做:
url(r'^reviews/', include(('reviews.urls', 'reviews'), namespace='reviews')),