In Django, how can I do a simple redirect directly from urls.py? Naturally I am a well organized guy, favoring the DRY principle, so I would like to get the target based on it's named url pattern, rather than hard coding the url.
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7 Answers
You can use RedirectView.
If you are on Django 1.4 or 1.5, you can do this:
from django.core.urlresolvers import reverse_lazy
from django.views.generic import RedirectView
urlpatterns = patterns('',
url(r'^some-page/$', RedirectView.as_view(url=reverse_lazy('my_named_pattern'), permanent=False)),
...
If you are on Django 1.6 or above, you can do this:
from django.views.generic import RedirectView
urlpatterns = patterns('',
url(r'^some-page/$', RedirectView.as_view(pattern_name='my_named_pattern', permanent=False)),
...
In Django 1.9, the default value of permanent has changed from True to False. Because of this, if you don't specify the permanent keyword argument, you may see this warning:
RemovedInDjango19Warning: Default value of 'RedirectView.permanent' will change from True to False in Django 1.9. Set an explicit value to silence this warning.
7 Comments
Daniel Backman
Just a note, remember that the RedirectView has permanent=True as default.
radtek
You can redirect everything! Like so:
(r'^.*/$', RedirectView.as_view(url='http://newurl.com')),
Gravity Grave
RedirectView will have permanent=False as default in Django 1.9.
Jay Modi
Can someone explain what is 'my_named_pattern' in above example.
tutuDajuju
passing
pattern_name argument to RedirectView executes reverse for you at call time using given pattern name. Other useful parameters include permanent and query_string. |
This works for me.
from django.views.generic import RedirectView
urlpatterns = patterns('',
url(r'^some-page/$', RedirectView.as_view(url='/')),
...
In above example '/' means it will redirect to index page,
where you can add any url patterns also.
1 Comment
Flimm
Just a note, remember that the
RedirectView has permanent=True as default in older versions of Django, and permanent=False as default in Django versions >= 1.9.for django v2+
from django.contrib import admin
from django.shortcuts import redirect
from django.urls import path, include
urlpatterns = [
# this example uses named URL 'hola-home' from app named hola
# for more redirect's usage options: https://docs.djangoproject.com/en/2.1/topics/http/shortcuts/
path('', lambda request: redirect('hola/', permanent=False)),
path('hola/', include("hola.urls")),
path('admin/', admin.site.urls),
]
1 Comment
Glushiator
@Ali Permanent redirects usually have HTTP code 301, temporary redirects usually have code 302. Permanent redirects are used to inform the browser of site's structural changes. Temporary redirects are used to indicate a new page to be shown after a dynamic server side action like login. According to the documentation
permanent=False is the default. For more details please read: docs.djangoproject.com/en/2.1/topics/http/shortcuts/#redirect developer.mozilla.org/en-US/docs/Web/HTTP/…I was trying to redirect all 404s to the home page and the following worked great:
from django.views.generic import RedirectView
# under urlpatterns, added:
url(r'^.*/$', RedirectView.as_view(url='/home/')),
url(r'^$', RedirectView.as_view(url='/home/')),
Comments
This way is supported in older versions of django if you cannot support RedirectView
In view.py
def url_redirect(request):
return HttpResponseRedirect("/new_url/")
In the url.py
url(r'^old_url/$', "website.views.url_redirect", name="url-redirect"),
You can make it permanent by using HttpResponsePermanentRedirect
Comments
You could do straight on the urls.py just doing something like:
url(r'^story/(?P<pk>\d+)/',
lambda request, pk: HttpResponsePermanentRedirect('/new_story/{pk}'.format(pk=pk)))
Just ensure that you have the new URL ready to receive the redirect!! Also, pay attention to the kind of redirect, in the example I'm using Permanent Redirect
Comments
If you need to capture part of the url including slashes and directly reuse that in the redirect, you can use re_path like this:
from django.shortcuts import redirect
from django.urls import re_path
re_path(r"admin/old_module/(.+)$", lambda req, rest: redirect(f"/admin/new_module/{rest}"))
Here rest is a positional argument containing the first capture group (.+). More arguments/capture groups could be added in the same manner. (see re_path)
