4

每当我尝试访问我网站上的“blog/”时,Django 都会给我一个 404 错误,但我已经定义了我想要的 URL,它们应该与之匹配。

主要网址.py:

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)),
)

博客.urls.py:

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

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

404页面:

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.

网站结构:

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'
)
4

1 回答 1

6

patterns需要一个前缀作为其第一个参数,后跟零个或多个参数。所以这:

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

inblog.urls.py应该是这样的:

urlpatterns = patterns('', url(r'^$', views.index, name='index'))  # now has a prefix as first argument

在当前状态下,patterns函数 inblog.urls.py将返回一个 empty pattern_list,这意味着它url(r'^blog/', include('blog.urls'))不会返回任何模式。

于 2014-04-07T00:48:19.157 回答