我有这些网址文件:
项目/ URL:
urlpatterns = [
url(r'^$', include(public)), <--- URL IN ERROR!!!!
url(r'^member/', include(mviews)),
url(r'^admin/', admin.site.urls),
url(r'^api-auth/', include('rest_framework.urls', namespace='rest_framework'))
public / url:
urlpatterns = [
url(r'^$', views.index),
url(r'^login/', views.login)
]
mviews / URL:
urlpatterns = [
url(r'^$', views.index),
url(r'^api/', include(router.urls)),
]
在第一个网址文件中,显示include(public)
的第一个网址错误输出。如何将其设置为公共是“主页”网址组?感谢。
答案 0 :(得分:0)
我觉得你在这里有错字,你的文件名是错的(url.py),应该是urls.py
如果上面的行成立,那么您可以通过以下
从公共模块中实现分组URLurl(r'^home/', include('public.urls')),
url(r'^member/', include(mviews)),
url(r'^admin/', admin.site.urls),
url(r'^api-auth/', include('rest_framework.urls', namespace='rest_framework'))