TypeError：url（）得到一个意外的关键字参数'name_space'

Question

I know it looks like a simple problem but I'm still too new for sorting it out myself, so :

learning_log/urls.py

from django.conf.urls import include, url
from django.contrib import admin

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

Previously I had the following error so I've removed the include function

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.

---

learing_logs/urls.py

"""Defines URL patterns for learning_logs."""

    from django.conf.urls import  url

    from . import views

    urlpatterns=[
        #Homepage
        url(r'^$',views.index,name='index'),

    ]

views.py

from django.shortcuts import render

    #Create your views here.

    def index(request):
        """The home page for Learning Log"""
        return render(request,'learning_logs/index.html')

I'm using Django 2.0 and Python 3.6.1

Could you please advise why I'm getting TypeError with name_space arg, is that related to Django version, many thanks in advance.

Answer 1

You should use include() to include urls patterns from another urls.py

url(r'', include('learning_logs.urls')),

There is no name_space argument with an underscore. The include() function accepts namespace . However, as the error message suggests, you should set app_name in the included urls.py instead of using namespace . You don't need to use namespace unless you are including the same urls multiple times.

from . import views
app_name = 'learning_logs'
urlpatterns=[
    #Homepage
    url(r'^$',views.index,name='index'),

]

Answer 2

The parameter you want is likely name not name_space

url(r'', 'learning_logs.urls', name='learning_logs')

(Django 2.0 url() Docs)