无法使URL配置生效

Question

我有一个应用程序,我想成为该网站的根.所以,我在我的主urls.py中包含了以下urlpattern,如下所示:

(r'^$', include('search.urls')),

在我的搜索中urls.py,我按以下方式设置它:

urlpatterns = patterns('search.views',
    (r'^$', 'index'),
    (r'^results/$', 'results'),
)

索引urlpattern工作,它进入索引视图,但结果urlpattern不起作用.我尝试访问结果时收到以下错误:

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

   1. ^books/
   2. ^$
   3. ^admin/
   4. ^static/(?P.*)$

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

编辑:主要urls.py:

urlpatterns = patterns('',
    (r'^books/', include('books.urls')),
    (r'^$', include('search.urls')),
    (r'^admin/', include(admin.site.urls)),
    (r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root': settings.MEDIA_ROOT}),
)

Answer 1

试试这个(把搜索结果包括在最后并删除$):

urlpatterns = patterns('',
    (r'^books/', include('books.urls')),

    (r'^admin/', include(admin.site.urls)),


    (r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root': settings.MEDIA_ROOT}),
    (r'^', include('search.urls')),
)

这样做的原因是,只有当您的路径与^ $匹配时才会包含搜索网址,这意味着只有/ url才会匹配此模式,因为$标记了模式的结尾.我的urlconf的最后一行说:将所有其他内容转发到搜索urlconf(注意缺少$).请注意,此行之后的任何内容都将永远不会匹配,因为此行与任何内容都匹配.