问题
I am pretty new to django but experienced in Python and java web programming with different frameworks. I have made myself a nice little django app, but I cant seem to make it match www.mysite.com as opposed to www.mysite.com/myapp.
I have defined urls and views in my urls.conf which is currently not decoupled from the app (don't mind that).
urlpatterns = patterns('myapp.views',
(r'^myapp/$', 'index'),
(r'^myapp/(?P<some_id>\d+)/global_stats/$', 'global_stats'),
(r'^myapp/(?P<some_id>\d+)/player/(?P<player_id>\d+)/$', 'player_stats'),
)
All this works like a charm. If someone goes to www.mysite.com/myapp they will hit my index view which causes a http redirect to my "correct" default url.
So, how can I add a pattern that will do the same as (r'^myapp/$', 'index') but without the /myapp--that is, www.mysite.com should suffice?
I would think this would be very basic stuff... I tried adding a line like:
(r'^$', 'index'),
however this throws me in a loop...
Hope you django gurus out there can clarify this for me!
回答1:
This sounds strange.
Your latest attempt should work, but what I normally do - put
urlpatterns = patterns('',
(r'^$', lambda r: HttpResponseRedirect('myapp/')),
...
)
This scales better when you start adding new apps.
回答2:
I know that this question was asked 2 years ago, but I've faced the same problem and found a solution:
In the project urls.py
:
urlpatterns = patterns('',
url(r'^', include('my_app.urls')), #NOTE: without $
)
In my_app.urls.py
:
urlpatterns = patterns('',
url(r'^$', 'my_app.views.home', name='home'),
url(r'^v1/$', 'my_app.views.v1', name='name_1'),
url(r'^v2/$', 'my_app.views.v2', name='name_2'),
url(r'^v3/$', 'my_app.views.v3', name='name_3'),
)
回答3:
As I didn't see any answer for django 2.0 I thought I'll provide one. you have to use ''
as the root url. Here is an example from django 2.0 docs
urlpatterns = [
path('', main_views.homepage),
path('help/', include('apps.help.urls')),
path('credit/', include(extra_patterns)),
]
回答4:
Just put an empty raw regular expression: r''
I tested here and it worked perfectly.
urlpatterns = patterns('',
url(r'', include('homepage.urls')),
url(r'^homepage/', include('homepage.urls')),
url(r'^admin/', include(admin.site.urls)),
)
Hope it help!
回答5:
I know the answer is late, but i had my fair share of hunting recently. This is what i tried with CBV.. in Project urls.py
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^$', include('app_name.urls', namespace='app_name')),
]
PS: It is always recommended to use namespace. Gives a good advantage later on.
In App urls.py
urlpatterns = [
url(r'^$', views.IndexPageView.as_view(), name='index'),
]
回答6:
I also wanted to have the root of my domain to directly point to a view of a sub app.
At first I created the sub app with this command:
python3 manage.py startapp main offer_finder/main
In your case it would be:
python3 manage.py startapp myApp project/somedirectory/myApp
make sure that the directory exists: mkdir -p project/somedirectory/myApp
This is my project structure: Project structure
In my case I have these directories:
offer_finder_project/offer_finder/main # sub app
offer_finder_project/offer_finder/ # main source directory
in offer_finder_project/offer_finder/urls.py
I have this code:
from django.contrib import admin
from django.urls import path, include
urlpatterns = [
path('admin/', admin.site.urls),
path('', include('offer_finder.main.urls')),
]
And in offer_finder_project/offer_finder/main/urls.py
I have this code:
from django.urls import path
from . import views
urlpatterns = [
path('', views.index),
]
And the offer_finder_project/offer_finder/main/views.py
file simply contains some test code.
from django.http import HttpResponse
def index(request):
return HttpResponse("TEST app index")
This way your requests to your root domain are directed to your sub app. Hopefully this helps someone. This code was tested with Django 2.1.4.
来源:https://stackoverflow.com/questions/7580220/django-urls-how-to-map-root-to-app