2012-07-31 84 views
1

我正在自定义django注册模块。到目前为止,我传递的URL像Django注册模块success_url没有返回

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

# Uncomment the next two lines to enable the admin: 
# from django.contrib import admin 
# admin.autodiscover() 

from django.contrib import admin 
admin.autodiscover() 
from django.conf.urls.defaults import * 
from django.views.generic.simple import direct_to_template 
import registration.backends.default.urls as regUrls 

from profile import UserRegistrationForm 
from registration.views import register 
import regbackend, views 
from accounts import profile 
urlpatterns = patterns('',  
    # (r'^conf/admin/(.*)', admin.site.root), 
    url(r'^register/$', register, {'backend': 'registration.backends.default.DefaultBackend','success_url':profile,'form_class': UserRegistrationForm}, name='registration_register'), 
    (r'^accounts/', include(regUrls)), 
    url('^profile/$', direct_to_template, {'template': 'profile.html'}, name="profile"), 


) 

当URL请求我的错误No module named django.views,它是不会去success_url

我想我在做错了urls.py但我看不到什么。请帮助我。

在此先感谢。

+0

请你能更新你的文章,包括你的完整的urls.py? – Alasdair 2012-07-31 09:33:07

+0

另外某种堆栈跟踪会有所帮助。 – 2012-07-31 10:19:28

回答

0

尝试改变这一行:

url(r'^register/$', register, {'backend': 'registration.backends.default.DefaultBackend','success_url':profile,'form_class': UserRegistrationForm}, name='registration_register'), 

要这样:

url(r'^register/$', 
    RegistrationView.as_view(
     form_class=UserRegistrationForm, 
     success_url='profile/',), 
    name='registration_register', 
    ), 

我没有在这里使用了一个名为URL,但它可以更容易把它与一个硬编码的工作都当你确定一切正常时,将它改为命名。