不能让django网址工作

I've got a simple django setup in which I have one app called 'lists'. I now want http://127.0.0.1:8000/lists to show this app. So I changed my main urls.py to the following:

我有一个简单的django设置，其中有一个名为'lists'的应用程序。我现在想要http://127.0.0.1:8000/lists来显示这个应用程序。所以我将主urls.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)),
)

and I changed the urls.py which resides in the lists-folder (my app is called lists) to the following:

我将位于lists-folder（我的app称为列表）的urls.py更改为以下内容：

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

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

As far as I know I am following the instructions in the django tutorial perfectly well, but when I visit http://127.0.0.1:8000/lists (without trailing slash) it gives me the following error:

据我所知，我完全按照django教程中的说明操作，但是当我访问http://127.0.0.1:8000/lists（没有斜杠）时，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.

and when I visit http://127.0.0.1:8000/lists/ (with a trailing slash) it gives me the following error:

当我访问http://127.0.0.1:8000/lists/（带斜杠）时，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:

    ^admin/

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

I don't understand why it doesn't search for ^lists/ anymore when I visit the url with the trailing slash. Does anybody know what I am doing wrong here?

我不明白为什么当我使用尾部斜杠访问url时它不再搜索^ lists /。有谁知道我在做错了什么？

All tips are welcome!

欢迎所有提示！

1 个解决方案

#1

You're missing the empty string at the start of your patterns in lists urls.py.

您在列表urls.py中的模式开头缺少空字符串。

Try this:

尝试这个：

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

The blank string is a view prefix that you can use to assist with the DRY principal. It is used to prefix your views path.

空白字符串是一个视图前缀，可用于协助DRY主体。它用于为视图路径添加前缀。

For example, (extending your example above):

例如，（扩展上面的示例）：

Rather than having:

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'),
)

You can use:

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'),
)

To have multiple view prefixes just segment your urlpatterns.

要有多个视图前缀，只需分割您的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'),
)

#1