Python Can'；无法让django URL正常工作

我有一个简单的django设置，其中有一个名为“列表”的应用程序。我现在想要

http://127.0.0.1:8000/lists

以显示此应用程序。因此，我将我的主URL.py更改为以下内容：

from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^lists/', include('lists.urls')),
    url(r'^admin/', include(admin.site.urls)),
)

from django.conf.urls import patterns, url
from lists import views

urlpatterns = patterns(
    url(r'^$', views.index, name='index')
)

我将列表文件夹（我的应用程序称为lists）中的url.py更改为：

from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    url(r'^lists/', include('lists.urls')),
    url(r'^admin/', include(admin.site.urls)),
)

from django.conf.urls import patterns, url
from lists import views

urlpatterns = patterns(
    url(r'^$', views.index, name='index')
)

据我所知，我完全按照中的说明进行操作，但是当我访问

http://127.0.0.1:8000/lists

（不带尾随斜杠）它给出以下错误：

Page not found (404) Request Method:    GET Request URL:    http://127.0.0.1:8000/lists

Using the URLconf defined in companyLists.urls, Django tried these URL patterns, in this order:

    ^lists/
    ^admin/

The current URL, lists, didn't match any of these.

Page not found (404) Request Method:    GET Request URL:    http://127.0.0.1:8000/lists/

Using the URLconf defined in companyLists.urls, Django tried these URL patterns, in this order:

    ^admin/

The current URL, lists/, didn't match any of these.

当我访问

http://127.0.0.1:8000/lists/

（带有一个尾随斜杠）它给出了以下错误：

Page not found (404) Request Method:    GET Request URL:    http://127.0.0.1:8000/lists

Using the URLconf defined in companyLists.urls, Django tried these URL patterns, in this order:

    ^lists/
    ^admin/

The current URL, lists, didn't match any of these.

Page not found (404) Request Method:    GET Request URL:    http://127.0.0.1:8000/lists/

Using the URLconf defined in companyLists.urls, Django tried these URL patterns, in this order:

    ^admin/

The current URL, lists/, didn't match any of these.

我不明白为什么当我访问带有尾随斜杠的url时，它不再搜索^lists/。有人知道我做错了什么吗

---

欢迎所有提示

列表

url.py

中的

模式的开头缺少空字符串

试试这个：

urlpatterns = patterns('',
    url(r'^$', views.index, name='index')
)

空白字符串是一个可用于帮助DRY委托人的字符串。它用于为视图路径添加前缀

例如，（扩展上面的示例）：

而不是：

urlpatterns = patterns('',
    url(r'^$', views.index, name='index'),
    url(r'^homepage$', views.homepage, name='index'),
    url(r'^lists$', views.lists, name='index'),
    url(r'^detail$', views.detail, name='index'),
)

您可以使用：

urlpatterns = patterns('views',
    url(r'^$', index, name='index'),
    url(r'^homepage$', homepage, name='index'),
    url(r'^lists$', lists, name='index'),
    url(r'^detail$', detail, name='index'),
)

只需分段您的
urlpatterns

urlpatterns = patterns('views',
    url(r'^$', index, name='index'),
    url(r'^homepage$', homepage, name='index'),
    url(r'^lists$', lists, name='index'),
    url(r'^detail$', detail, name='index'),
)

urlpatterns += patterns('more_views',
    url(r'^extra_page$', extra_page, name='index'),
    url(r'^more_stuff$', something_else, name='index'),
)

啊，我忽略了。这确实有效！但最重要的是；为什么？为什么一开始就需要一个空字符串？现在更新了我的答案@kramer65-希望能让它更清楚。