TypeError: url() got an unexpected keyword argument 'name_space'

Question

我知道这看起来是个简单的问题，但我还是个新手，无法自己解决，所以:

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'),
        ]

之前我有以下错误，所以我删除了包含函数

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')

我正在使用 Django 2.0 和 Python 3.6.1

你能告诉我为什么我得到 TypeError with name_space arg，这与 Django 版本有关，非常感谢。

Answer 1

你想要的参数很可能是name而不是name_space

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

(Django 2.0 url() Docs)

Answer 2

您应该使用 include() 来包含来自另一个 urls.py

的 url 模式

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

没有带下划线的 name_space 参数。 include() 函数接受 namespace。但是，正如错误消息所建议的那样，您应该在包含的 urls.py 中设置 app_name 而不是使用 namespace。您不需要使用 namespace 除非您多次包含相同的网址。

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

]

TypeError: url() got an unexpected keyword argument 'name_space'

TypeError: url() got an unexpected keyword argument 'name_space'

python

django

django-urls

typeerror