I use Django 1.5 + django-registration 0.9...
How to make email field unique in model User?
from registration.forms import RegistrationFormUniqueEmail
url(r'^accounts/register/$', 'registration.views.register',
{'form_class': RegistrationFormUniqueEmail,
'backend': 'registration.backends.default.DefaultBackend'},
name='registration_register'),
This solution is not suitable
Could not import registration.views.register. View does not exist in module registration.views.
Many thanks, #Alasdair
How to use different view for django-registration?
urls.py:
from registration.backends.default.views import RegistrationView
from registration.forms import RegistrationFormUniqueEmail
class RegistrationViewUniqueEmail(RegistrationView):
form_class = RegistrationFormUniqueEmail
urlpatterns = patterns('',
....
url(r'^user/register', RegistrationViewUniqueEmail.as_view(),
name='registration_register'),
...
Another approach is to pass the form class into the as_view() method directly as in the following example:
url(r'^user/register/$', RegistrationView.as_view(
form_class=RegistrationFormUniqueEmail),
name='registration_register'),
A complete urls.py for this:
from django.conf.urls import patterns, include, url
from registration.forms import RegistrationFormUniqueEmail
from registration.backends.default.views import RegistrationView
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
# Examples:
# url(r'^$', 'django_registration_demo.views.home', name='home'),
# url(r'^blog/', include('blog.urls')),
url(r'^admin/', include(admin.site.urls)),
url(r'^accounts/', include('django.contrib.auth.urls')),
# enable unique email registration feature
url(r'^accounts/register/$',
RegistrationView.as_view(form_class=RegistrationFormUniqueEmail),
name='registration_register'),
url(r'^accounts/', include('registration.backends.default.urls'))
)
I've also create a complete demo for django-registration, which enables unique email feature , see:
https://github.com/xiaohanyu/django-registration-demo
Related
Say i have a django project which have 4 apps 1 app is for logging in
project urls:
from django.conf.urls import url, include
from django.contrib import admin
from django.conf import settings
urlpatterns = [
url(r'^', include('Login.urls')),
url(r'^admin/', admin.site.urls),
]
settings:
LOGIN_REDIRECT_URL = '/'
login app urls:
from django.conf.urls import url
from django.contrib.auth import views as auth_views
urlpatterns = [
url(r'^login/$', auth_views.login, name='login'),
url(r'^logout/$', auth_views.logout, {'next_page': '/'}, name='logout'),
]
now say i have 3 apps based on user type i.e. Admin, Team Lead, Worker
so i want to redirect user based on their type of employment.
how can i achieve this ?
any help is appreciated
I guess that you will manually add/edit the user groups.
That can be done from django admin panel.
Here is the group/permission django docs: Permissions and Authorization
This is what I did:
I've created an index view in order to get the user group and redirect to the respective view that you want.
So, index view in views.py:
from django.contrib.auth.decorators import login_required
#login_required
def index(request):
group = request.user.groups.filter(user=request.user)[0]
if group.name=="employees":
return HttpResponseRedirect(reverse('worker'))
elif group.name=="teamLeader":
return HttpResponseRedirect(reverse('teamLeader'))
elif group.name=="admin":
return HttpResponseRedirect(reverse('adm'))
context = {}
template = "index.html"
return render(request, template, context)
And urls in urls.py:
from django.conf.urls import url
from django.contrib import admin
from django.contrib.auth import views as auth_views
from app import views as app_views
urlpatterns = [
url(r'^login/$', auth_views.login, name='login'),
url(r'^$', app_views.index, name='index'),
url(r'^employees/$', app_views.employees, name='employees'),
url(r'^teamLeader/$', app_views.teamLeader, name='teamLeader'),
url(r'^adm/$', app_views.adm, name='adm'),
]
In settings.py
LOGIN_URL = '/login'
LOGIN_REDIRECT_URL = '/'
Is this what you want to do?
I'm trying to customize the url of entries in zinnia to show slugs of entries only, ie .../blog/slug.
I've been following closely the documentation here - I've overwritten the get_absolute_url method, I've added the view and configured the urls and registered the _base model in django settings - yet the error persists:
zinnia_customized models.py:
from django.db import models
from zinnia.models_bases.entry import AbstractEntry
class EntryWithNewUrl(AbstractEntry):
"""Entry with '/blog/<id>/' URL"""
#models.permalink
def get_absolute_url(self):
return ('zinnia:entry_detail', (),
{'slug': self.slug})
class Meta(AbstractEntry.Meta):
abstract = True
zinnia_customized views.py:
from django.views.generic.detail import DetailView
from zinnia.models.entry import Entry
from zinnia.views.mixins.entry_preview import EntryPreviewMixin
from zinnia.views.mixins.entry_protection import EntryProtectionMixin
class EntryDetail(EntryPreviewMixin, EntryProtectionMixin, DetailView):
queryset = Entry.published.on_site()
template_name_field = 'template'
project urls.py:
urlpatterns = [
url(r'^$', TemplateView.as_view(template_name='pages/home.html'), name='home'),
url(r'^about/$', TemplateView.as_view(template_name='pages/about.html'), name='about'),
url(r'^admin/tools/', include('admin_tools.urls')),
url(settings.ADMIN_URL, include(admin.site.urls)),
url(r'^users/', include('anpene.users.urls', namespace='users')),
url(r'^accounts/', include('allauth.urls')),
url(r'^blog/', include('zinnia_customized.urls', namespace='zinnia')),
url(r'^comments/', include('django_comments.urls')),
]
zinnia_customized urls.py:
blog_urls = [
url(r'^', include('zinnia.urls.capabilities')),
url(r'^search/', include('zinnia.urls.search')),
url(r'^sitemap/', include('zinnia.urls.sitemap')),
url(r'^trackback/', include('zinnia.urls.trackback')),
url(r'^blog/tags/', include('zinnia.urls.tags')),
url(r'^blog/feeds/', include('zinnia.urls.feeds')),
url(r'^blog/authors/', include('zinnia.urls.authors')),
url(r'^blog/categories/', include('zinnia.urls.categories')),
# url(r'^blog/', include('zinnia.urls.entries')),
url(r'^blog/', include('zinnia_customized.urls.entries')),
url(r'^blog/', include('zinnia.urls.archives')),
url(r'^blog/', include('zinnia.urls.shortlink')),
url(r'^blog/', include('zinnia.urls.quick_entry')),
]
urlpatterns += patterns('',
url(r'^', include(blog_urls), name='blog')
)
zinnia_customized app urls/entries.py:
from django.conf.urls import url
from django.conf.urls import patterns
from zinnia_customized.views import EntryDetail
urlpatterns = [
url(r'^(?P<slug>[\w-]+)/$', EntryDetail.as_view(), name='entry_detail'),
]
zinnia_customized admin.py:
from django.contrib import admin
from django.utils.translation import ugettext_lazy as _
from zinnia.models.entry import Entry
from zinnia.admin.entry import EntryAdmin
class EntryUrlAdmin(EntryAdmin):
"""blank"""
admin.site.register(Entry, EntryUrlAdmin)
settings:
...
ZINNIA_ENTRY_BASE_MODEL = 'zinnia_customized.models.EntryWithNewUrl'
...
And the error:
NoReverseMatch at /blog/
Reverse for 'entry_detail' with arguments '()' and keyword arguments '{'slug': u'pies'}' not found. 1 pattern(s) tried: [u'blog/(?P<year>\\d{4})/(?P<month>\\d{2})/(?P<day>\\d{2})/(?P<slug>[-\\w]+)/$']
My problem was that I've created a folder called urls in my zinnia_customized, therefore django wasn't sure if it was supposed to use the folder or urls.py
I am new to django and python. During url mapping to views i am getting following error:
TypeError: view must be a callable or a list/tuple in the case of include().
Urls. py code:-
from django.conf.urls import url
from django.contrib import admin
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^posts/$', "posts.views.post_home"), #posts is module and post_home
] # is a function in view.
views.py code:-
from django.shortcuts import render
from django.http import HttpResponse
# Create your views here.
#function based views
def post_home(request):
response = "<h1>Success</h1>"
return HttpResponse(response)
Traceback
In 1.10, you can no longer pass import paths to url(), you need to pass the actual view function:
from posts.views import post_home
urlpatterns = [
...
url(r'^posts/$', post_home),
]
Replace your admin url pattern with this
url(r'^admin/', include(admin.site.urls))
So your urls.py becomes :
from django.conf.urls import url, include
from django.contrib import admin
urlpatterns = [
url(r'^admin/', include(admin.site.urls)),
url(r'^posts/$', "posts.views.post_home"), #posts is module and post_home
]
admin urls are callable by include (before 1.9).
For Django 1.11.2
In the main urls.py write :
from django.conf.urls import include,url
from django.contrib import admin
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^posts/', include("Post.urls")),
]
And in the appname/urls.py file write:
from django.conf.urls import url
from . import views
urlpatterns = [
url(r'^$',views.post_home),
]
Answer is in project-dir/urls.py
Including another URLconf
1. Import the include() function: from django.conf.urls import url, include
2. Add a URL to urlpatterns: url(r'^blog/', include('blog.urls'))
Just to complement the answer from #knbk, we could use the template below:
as is in 1.9:
from django.conf.urls import url, include
urlpatterns = [
url(r'^admin/', admin.site.urls), #it's not allowed to use the include() in the admin.urls
url(r'^posts/$', include(posts.views.post_home),
]
as should be in 1.10:
from your_project_django.your_app_django.view import name_of_your_view
urlpatterns = [
...
url(r'^name_of_the_view/$', name_of_the_view),
]
Remember to create in your_app_django >> views.py the function to render your view.
You need to pass actual view function
from posts.views import post_home
urlpatterns = [
...
url(r'^posts/$', post_home),
]
This works fine!
You can have a read at URL Dispatcher Django
and here Common Reguler Expressions Django URLs
I just installed userena, and had the example working from the tutorial, but as soon as I added in a single line in URLS.py, I'm getting an error. In the example below, I added the line mapping the home function from views.py
Now the issue I'm having is that when I go to 127.0.0.1/8000, I get TypeError: string is not callable, but then oddly, if I go to accounts/signup or accounts/signin, I am getting the template that should be appearing if i go to 127.0.0.1/8000.
from django.conf import settings
from django.conf.urls import patterns, include, url
from django.conf.urls.static import static
from django.views.generic import TemplateView
from accounts import views
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
url(r"^$", 'home'),
url(r'^admin/', include(admin.site.urls)),
(r'^accounts/', include('userena.urls')),
)
Here is my accounts/views.py
from django.shortcuts import render
from django.http import HttpResponseRedirect
def home(request):
return render('homepage.html')
You need to remove the quotes in the url and import that view
from accounts.views import home
urlpatterns = patterns('',
url(r"^$", home),
url(r'^admin/', include(admin.site.urls)),
(r'^accounts/', include('userena.urls')),
)
You can steel use the strings in the url() but you must use the format 'app.views.viewname'
urlpatterns = patterns('',
url(r"^$", 'accounts.views.home'),
url(r'^admin/', include(admin.site.urls)),
(r'^accounts/', include('userena.urls')),
)
Or name the module in the first argument as string to patterns()
urlpatterns = patterns('accounts.views',
url(r"^$", 'home'),
url(r'^admin/', include(admin.site.urls)),
(r'^accounts/', include('userena.urls')),
)
the issue was I forgot to include the request in the return render.
The correct answer is that render is being called incorrectly. Actually, the views.py file would raise a SyntaxError, but we'll let that slide :)
# views.py
from django.shortcuts import render
def home(request):
return render(request, 'homepage.html')
The code is
#urls.py
from django.conf.urls import patterns, url
from employees import views
from schdeules import views
urlpatterns = patterns('',
url(r'^$', views.home, name='home'),
url(r'^accounts/logout/$', 'django.contrib.auth.views.logout'),
url(r'^accounts/login/$', 'django.contrib.auth.views.login', {'template_name': 'admin/login.html'}),
url(r'^accounts/$', 'django.views.generic.simple.redirect_to', {'url': '/'}),
url(r'^accounts/profile/$', 'django.views.generic.simple.redirect_to', {'url': '/'}),
)
#views.py
# Create your views here.
from django.contrib.auth.decorators import login_required
from django.shortcuts import render
#login_required
def home(request):
welecome = 'hai welcome to opas'
context = {'temp_var': welecome}
return render(request, 'schdeules/home.html')
and iam getting an error
Exception Type: ViewDoesNotExist at /opas/
Exception Value: Could not import django.views.generic.simple.redirect_to. Parent module django.views.generic.simple does not exist.
i want to use django default login and logout modules.
if successfully logged in then i want to redirect to home page.
Thanks in advance.
The function-based generic view redirect_to was deprecated in Django 1.3, and removed in Django 1.5. Use the class-based generic view RedirectView instead.
from django.views.generic.base import RedirectView
urlpaterns = ('',
...
url(r'^accounts/$', RedirectView.as_view(url='/')),
url(r'^accounts/profile/$', RedirectView.as_view(url='/')),
)
Note that you don't have to include url patterns for /accounts/ and /accounts/profile/. You may be better to set LOGIN_REDIRECT_VIEW in your settings, so that users are redirected straight to the home page after logging in.
LOGIN_REDIRECT_VIEW = 'home' # using a named url pattern requires Django 1.5 or later