Django URL提供404

Question

每当我试图访问我的站点上的“blog/”时，Django就会给我一个404错误，但是我已经定义了我想要的url，它们应该与之匹配。在

主网址.py公司名称：

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

from django.contrib import admin
admin.autodiscover()
from blog import views

urlpatterns = patterns('',
    # Examples:
    # url(r'^$', 'mySiteProject.views.home', name='home'),
    # url(r'^blog/', include('blog.urls')),

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

在博客.url.py公司名称：

^{pr2}$

404页码：

Page not found (404)
Request Method:     GET
Request URL:    http://localhost:8000/blog/

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

    ^admin/

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

场地结构：

mySiteProject
    blog
        admin.py
        models.py
        tests.py
        views.py
        urls.py
        __init__.py
    mySiteProject
        wsgi.py
        settings.py
        urls.py
        __init__.py
    manage.py
    db.sqlite3

已安装的应用程序：

INSTALLED_APPS = (
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.messages',
    'django.contrib.staticfiles',

    'blog'
)

Answer 1

patterns需要前缀作为其第一个参数，后跟零个或多个参数。所以这个：

urlpatterns = patterns(url(r'^$',views.index,name='index'))  # won't work

在blog.urls.py中应该如下所示：

^{pr2}$

在当前状态下，blog.urls.py中的patterns函数将返回一个空的pattern_list，这意味着url(r'^blog/', include('blog.urls'))将不返回任何模式。在