Django无法恢复网址,即使提供了预期的kwarg。
这是root urls.py:
from django.conf import settings
from django.conf.urls import include, url
from django.conf.urls.static import static
from django.contrib import admin
urlpatterns = [
url(r'^admin/', include(admin.site.urls)),
url(r'^media/(?P<path>.*)$','django.views.static.serve',{'document_root': settings.MEDIA_ROOT}),
url(r'^ckeditor/', include('ckeditor_uploader.urls')),
url(r'^static/(?P<path>.*)$','django.views.static.serve',{'document_root': settings.STATIC_ROOT}),
url(r'^(?P<domain>\w+)', include('frontend.urls')),
] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
这是frontend urls.py: 来自django.conf.urls import include,patterns,url
from . import views
from .views import MyprofileView
from .views import SuccessView
from .views import CompanyView
from .views import SubscriptionView
from django.views.decorators.csrf import csrf_exempt
urlpatterns = patterns('',
url(r'/success(/?)$', SuccessView.as_view(), name='success'),
url(r'/subscribe(/?)$', SubscriptionView.as_view(), name='subscribe'),
url(r'^(/?)$', MyprofileView.as_view(), name='home'),
url(r'/api/v1/', include('cpprofile.api.urls')),
url(r'/product', include('product_information.urls')),
url(r'/corporations/(?P<company>\d+)$', CompanyView.as_view(), name='company_page'),
url(r'^/(?P<subscription>\w+)/product/pay/return/(?P<amount>\d+)/(?P<currency>\w+)/(?P<id>\d+)?$',
views.payment_return, name='subscription_product_payment_return'),
)
以下是我试图在view.py MyprofileView中调用它的方法:
context['subscribe_url'] = redirect('subscribe', kwargs={'domain': 'up'})
这里可能有什么问题?
由于
以下是我遇到的错误:
django.core.urlresolvers.NoReverseMatch
NoReverseMatch: Reverse for 'subscribe' with arguments '()' and keyword arguments '{'domain': 'up'}' not found. 1 pattern(s) tried: ['(?P<domain>\\w+)/subscribe(/?)$']
答案 0 :(得分:2)
您必须解压缩kwargs
。
<强>解决方案强>:
kwargs = {'domain': 'up'}
redirect('app_name:subscribe', **kwargs)
编辑:这将有效,无需更改网址 EDIT2:将应用程序的名称和冒号添加到网址名称。这会在app命名空间中找到url。
答案 1 :(得分:1)
我怀疑这是因为<div class="full-page login-page" data-color="blue" data-image="./assets/img/hola.jpg">
。这会捕获(/?)
或''
。因此,您必须将其作为非关键字参数传递:
'/'
所以这是Sachin Kukreja所说的。
答案 2 :(得分:1)
您需要使用reverse来获取正确的URL,然后重定向到该URL。
from django.core.urlresolvers import reverse
return redirect(reverse('subscribe', kwargs={'domain': 'up'}))
在您的情况下,您似乎尝试将url分配给上下文变量,您根本不应该使用重定向。反向解析URL,重定向返回响应。
context['subscribe_url'] = reverse('subscribe', kwargs={'domain': 'up'})
也可能希望在urlconf中遵循最佳实践以保持一致性,即使用'/'结束所有模式,但不要以'/'开头。正如你在root配置中为大多数人所做的那样:
url(r'^admin/', include(admin.site.urls)), <-- good