I have a basic splash page, and I am trying to redirect all urls to the splash EXCEPT for the thank you page (which is linked to after the email form is submitted).
How do I make it such that all my urls will redirect to the splash page with the exception of this one page? Currently, ALL of my urls are re-directing, even the exception. Here is my code:
urlpatterns = patterns('',
(r'^$', 'index'),
(r'^thanks/$', 'thanks'),
(r'^', 'index_redirect'),
Thank you.
In Django 1.3 you can use the redirect_to along with a pattern that matches everything.
from django.views.generic.simple import redirect_to
urlpatterns = patterns('',
(r'^$', 'index'),
(r'^thanks/$', 'thanks'),
(r'^.*$', redirect_to, {'url': '/'}),
)
WARNING: this WILL match your static resources and images etc.
Related
I am trying to find a way to display my django login page when the user ask for an unwanted url? Which syntax should I user ?
As of today I have
from django.conf.urls import patterns, include, url
from django.contrib import admin
urlpatterns = patterns('',
# Examples:
url(r'^login' , 'database.views.index', name='login'),
url(r'^create-user/' , 'database.views.account_creation', name='create_user'),
url(r'^get-details/' , 'database.views.get_details', name='get-details'),
url(r'^upload-csv' , 'database.views.upload_csv', name='upload_csv'),
# url(r'^blog/', include('blog.urls')),
url(r'^admin/', include(admin.site.urls)),
#url(r'^' , 'database.views.index', name='login'),
)
I would like that if a user ask for a crazy url, it would be directed to the login url (ie view.index function).
Any idea ?
Without commenting on whether you should do this, Django will attempt to match your url patterns in order. So if you want a fall-through / catch-all handler, put this last:
url(r'^.*', 'database.views.index', name='unmatched')
Wondering if I could get some help with a second pair of eyes. Started working Django about two months ago learning in my spare time. Long story short: My index and post view work fine. I recently created a contact html\template and view which looks like it goes right back to my index page? Here is my project urls.py
from django.conf.urls import patterns, include, url
from django.contrib import admin
from .settings import MEDIA_ROOT
urlpatterns = patterns('',
url(r'^admin/', include(admin.site.urls)),
url(r'^blog/', include('blog.urls')),
url(r'^contact/', include('blog.urls')),
url(r'^media/(?P<path>.*)$', 'django.views.static.serve',
{'document_root': MEDIA_ROOT}),
)
Here is my app urls.py
from django.conf.urls import patterns, include, url
from blog import views
urlpatterns = patterns('blog.views',
url(r'^(?P<post_name>\w+)/$', views.post, name='post'),
url(r'^contact/$', views.contact, name='contact'),
url(r'^$', views.index, name='index'),
)
If I change my contact view to come up as my default the view it comes up fine.
urlpatterns = patterns('blog.views',
url(r'^(?P<post_name>\w+)/$', views.post, name='post'),
url(r'^$', views.contact, name='contact'),
#url(r'^$', views.index, name='index'),
)
As soon as switch it back to the original state when I click on the contact link on my page it goes right to my index page. I have to be doing something wrong with the URL dispatcher part but I'm not sure what.
Your contact urlpattern is not matching the URL "/contact", it is matching "/blog/contact" and "/contact/contact". Since you've included the blogs urlpatterns under both those prefixes, the thing that matches "/contact" is the URL without anything after the prefix, ie the index.
You probably don't want to put the contact pattern into the included file at all: just match it directly in the base project file.
Please mind the order of the urls in your app's urls.py. Try to put your views.contact view before your views.post view. Keep in mind: Django chooses the first matching regular expression, which is your views.post view.
I ran into a similar problem and described the solution here.
Here your contact view has the url /contact/contact/ and index view has the url /contact/
So you have rewrite the project urls as
from django.conf.urls import patterns, include, url
from django.contrib import admin
from .settings import MEDIA_ROOT
urlpatterns = patterns('',
url(r'^admin/', include(admin.site.urls)),
url(r'^blog/', include('blog.urls')),
url(r'^', include('blog.urls')),
url(r'^media/(?P<path>.*)$', 'django.views.static.serve',
{'document_root': MEDIA_ROOT}),
)
i am learning django by using django 1.6 documents tutorial 1 - 6.
this round is my 4th try and, previous 3 try was successful and i understand more on every try.
i am in tutorial 3 now, to create views.
according to the documents, after i created a view, i need to map it to a URL.
so i follow the documents to add a urls.py in the polls directory.
and then i follow the document to add include() to mysite/urls.py
i am able to so called wired an index view into the polls view.
now if i go back to localhost:8000, i get an error page,
so my question is
1) WHY?
2) how to get back my localhost:8080 index page co-exist with the polls index page?
thank you very much everyone for your time. i am new and sorry for so simple question.
thank you.
updated:
this is my urls.py
from django.conf.urls import patterns, include, url
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
# Examples:
# url(r'^$', 'mysite.views.home', name='home'),
# url(r'^blog/', include('blog.urls')),
url(r'^admin/', include(admin.site.urls)),
url(r'^polls/', include('polls.urls')),
)
this is my polls/urls.py
from django.conf.urls import patterns, url
from polls import views
urlpatterns = patterns('',
url(r'^$', views.index, name='index')
)
the error msg is:
Using the URLconf defined in mysite.urls, Django tried these URL patterns, in this order:
^admin/
^polls/
The current URL, , didn't match any of these.
http://localhost:8000 index page will be displayed when you created a project (mysite). After creating an application and including it in installed apps in settings.py, you will no longer view the index page of the project, rather your admin page of the application (polls) will be visible at http://localhost:8000/admin
And if you had created any views with the following patterns in polls/urls.py
urlpatterns = patterns('',
url(r'^$', views.index, name='index'),
url(r'^blog/$', views.blog, name='blog'),
)
You need to visit http://localhost:8000/polls for your index page and http://localhost:8000/polls/blogs for your poll blogs page
Editing my answer for step by step approach:
create index.html at polls/templates/polls directory.
in polls/views.py
from django.shortcuts import render_to_response
def pollindex(request):
return render_to_response('polls/index.html', context_instance=RequestContext(request))
in polls/urls.py
urlpatterns = patterns('',
url(r'^index/$', views.pollsindex, name='index'),
)
in your project/urls.py
urlpatterns = patterns('',
url(r'^polls/', include('polls.urls')),
url(r'^admin/', include(admin.site.urls)),
)
And now you can view your template index.html at http://localhost:8000/polls/index
I have a url setup with the following view (the url is in the app and the app urls are included in the project):
url(r'^details/(?P<outage_id>\d+)/$', 'outage.views.outage_details'),
def outage_details(request, outage_id=1):
outage_info = Outages.objects.filter(id=outage_id)
return render_to_response('templates/outage/details.html', {'outage_info': outage_info}, context_instance=RequestContext(request))
When I click on the link from http://localhost:8000 the url in the browser changes to http://localhost:8000/outage/details/1 as it should, but the view doesn't render the right template. The page stays the same. I don't get any errors, the url changes in the browser but the details.html template doesn't render. There is an outage in the DB with an ID of 1.
Any thoughts?
The regular expression r'^details/(?P<outage_id>\d+)/$' does not match the URL http://localhost:8000/outage/details/1. However, it should match the expression r'^outage/details/(?P<outage_id>\d+)/$'.
Perhaps, you can post your entire urls.py to find out which view is actually being called, since you don't get any errors. I suspect your home page is being called for all URLs.
Here is my url setup:
project/urls.py
urlpatterns = patterns('',
url(r'^$', 'outage.views.show_outages'),
url(r'^inventory/', include('inventory.urls')),
url(r'^outage/', include('outage.urls')),
url(r'^login', 'django.contrib.auth.views.login', {'template_name': 'templates/auth/login.html'}),
url(r'^logout', 'django.contrib.auth.views.logout', {'next_page': '/'}),
url(r'^admin/', include(admin.site.urls)),
)
outage/urls.py
urlpatterns = patterns('',
url(r'^', 'outage.views.show_outages'),
url(r'^notes/(?P<outage_id>\d+)/$', 'outage.views.outage_notes', name='notes'),
)
I have since changed the details to notes, since I had another page in a different app with a details url and I didn't want it somehow confusing things.
I am using Django 1.4 for my project Builder
If I get a request like http://xxx.xx.xx:8000/index.html, how can I redirect to /home/?
Redirect all /index.html requests to /home/ using generic.simple.redirect_to
from django.views.generic.simple import redirect_to
urlpatterns = patterns('',
(r'^index.html$', redirect_to, {'url': '/home/'}),
(r'^/home/$', 'myapp.views.my_home_view'),
)
You need to add an url to your urlconf
urlpatterns = patterns('',
...
(r'^index\.html$', 'myapp.views.my_home_view'),
...
)
EDIT
An alternative option is to make use of the django.contrib.redirects app, and add an entry via the admin that redirects from index.html to home/. This is quicker and more configurable
Use redirect_to method of generic.simple:
from django.views.generic.simple import redirect_to
urlpatterns = patterns('',
(r'^home/$', 'yourapp.views.home', name='home'),
(r'^index.html$', redirect_to, {'url': '/home/'}),
)