无法导入名称模式

Question

在我写信之前urls.py,我的代码......一切都很完美.现在我有问题 - 不能去我的网站."无法导入名称模式"

我urls.py是:

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

他们说这里有什么错误.

Answer 1

从Django 1.10开始,该patterns模块已被删除(自1.8以来已被弃用).

幸运的是,它应该是一个简单的编辑,以删除有问题的代码,因为urlpatterns 现在应该存储在一个普通的列表中:

urlpatterns = [
    url(r'^admin/', include(admin.site.urls)),
    # ... your url patterns
]

Answer 2

你不需要那些进口.你在urls.py(开始)中唯一需要的是:

from django.conf.urls.defaults import *

# This two if you want to enable the Django Admin: (recommended)
from django.contrib import admin
admin.autodiscover()

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

注意:此解决方案适用于Django <1.6.这实际上是Django自己生成的代码.对于较新版本,请参阅Jacob Hume的回答.

Answer 3

是:

from django.conf.urls.defaults import ... # is for django 1.3
from django.conf.urls  import ...         # is for django 1.4

我也遇到了这个问题.

Answer 4

模块模块不受支持..我使用过这个.

from django.conf.urls import *
from django.contrib import admin
admin.autodiscover()

urlpatterns = [
    url(r'^admin/', include(admin.site.urls)),
    # ... your url patterns
]

Answer 5

这是适合我的代码.我的django版本是1.10.4决赛

from django.conf.urls import url, include

from django.contrib import admin
admin.autodiscover()

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

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