Develop Reference

redirection from old urls to new urls

Hello everyone I want to redirect URLs like this
from
https://www.example.co.uk/pages/terms-conditions
to
https://www.example.co.uk/terms-conditions
I developed the updated site using the Django framework and now want to redirect old site URLs to new URLs and not losing traffic. what is the best way of doing…?
New site is not published it is on local machine
my view.py
my urls.py

I guess this is not best practice but you can just redirect the user to the right url if he hits the other one in your views.py like:
def old_view(request):
return redirect('https://www.example.co.uk/terms-conditions')
def new_view(request):
return render(request, "your_app/conditions.html")
with urls.py like:
path('pages/terms-conditions/', views.old_view, name='old_view')
path('terms-conditions/', views.new_view, name='new_view')

Assuming you want to remove the application name from the urls. In your project's urls.py NOT your applications urls.py. You should have a line similar to the following:
path('pages/', include('pages.urls', namespace='pages')),
replace the 'pages/' with just ''.
Note: Using the pages/ allows you to provide segregation between your applications to ensure the urls do not conflict

How to override a Django url path that is provided by a library?

I'm currently building out an authentication server using Django, djangorestframework, and django-rest-auth. My problem is actually pretty simple I think, but I haven't really been able to find any resources on it.
Here's my issue, in django-rest-auth there is a specific url to change a user's password, which is /rest-auth/password/reset/. I'd like the url to instead be /auth/password/change/, but don't want to edit the library code to do so.
The issue is that as of now, in my url.py file, I have the rest-auth urls imported as such:
from django.urls import path, include
urlpatterns = [
path('', include('rest_auth.urls')),
]
So it just imports the urls as written in the library. How can I change the specific url to what I want?

You just need to add a url like this
path('/auth/password/change/', your_view)
your_view will be same as view of /rest-auth/password/reset/ which is PasswordResetView.

Modifying and creating sites in djangos admin app

I have to include multiple changes to djangos admin panel, so I decided to fork the django admin app into my own django project.
As I was working with this admin app I recognized, that the site registration and template handling differs from the apps, that are normally created in django.
For instance, I want to keep the old admin index.html template and view, for backup and safety reasons but the landing page should be replaced by a custom page.
For that of course I need to change admin/templates/index.html and /admin/sites.py respectively.
I copied the old index function in admin/sites.py to old_index.py and created a old_index.html in the template folder.
But if I try to reference to old_index.html in my new index.html with
old index
I got an NoReverseMatch-Exception thrown. Unfortunately I did not found more information about how the django admin app itself register new views and sites, so an example or description would be helpful.
Creating separate views for the admin app in the distinct other apps in my project is no real option, due the high amount of changes, that need to be done.
The main urls.py
from django.conf.urls import patterns, include, url
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
# Examples:
# url(r'^$', 'django_project.views.home', name='home'),
url(r'^polls/', include('other_app.urls', namespace="other_app")),
url(r'^admin/', include(admin.site.urls)),
)
The admin app itself does not provide a urls.py file and the views.py is exactely the same as in django.contrib.admin I just copied the function index to a new function called old_index, referencing to a template old_index.html.
Maybe the point did not get so clear, as I expected. I copied the whole admin app in my project and want to add a custom defined site to it, regardless where. But I failed to understand how sites and views are registered in the admin app itself, because the way is different from the custom apps you create normally in django.
So, is it possible (and how) to add a custom site in the django.contrib.admin app?

I think you need to create your own AdminSite for custom purposes and keep default as it is. More about this you can find here: https://docs.djangoproject.com/en/1.6/ref/contrib/admin/#adminsite-objects and here https://docs.djangoproject.com/en/dev/ref/contrib/admin/#multiple-admin-sites-in-the-same-urlconf
Update:
You need to edit get_urls method of AdminSite class - add:
url(r'^$', wrap(self.old_index), name='old_index')
to urlpatterns variable. And rename old index method to old_index.

Django url parameter passing

I am not even sure what category in Django this question falls in and I am very new to django. I have tried looking for Django post requests, parameter passing and even checked under Django APIs but have not found what I am looking for. What I am trying to do is create an API for my client but it must be done in Django. If I was to do this in .Net I could use http post and http get and web services but I am not at all sure how this is done in Django. What my client wants to do is to be able to see:
Enter username and password in url with parameters for date and id and be able to view rooms available based on what is entered
Enter username, password and dates and room code to be able to book the room
No interface needed just simple parameter passing through url. Is this possible with Django and if yes can somebody please point me in the right direction.

What you're looking for are captured parameters
Below is a code snippet from the above link.
# urls.py
from django.conf.urls import patterns, url
urlpatterns = patterns('blog.views',
url(r'^blog/(?P<year>\d{4})/$', 'year_archive', {'foo': 'bar'}),
)
# views.py
def year_archive(request, year, foo=None):
# view method definition

As of Django 1.10, patterns was removed from Django.
Here is a 2019 Django 2.2 solution.
It looks like re_path is the current recommended tool for capturing parameters via regular expressions:
https://docs.djangoproject.com/en/2.2/ref/urls/#re-path
# urls.py
from django.urls import re_path
from myapp import views
urlpatterns = [
re_path(r'^blog/(?P<year>\d{4})/$', 'year_archive', {'foo': 'bar'}),
]
# views.py
def year_archive(request, year, foo=None):
# view method definition

Customized views with django-registration

I need to make a very simple modification -- require that certain views only show up when a user is not authenticated -- to django-registration default views. For example, if I am logged in, I don't want users to be able to visit the /register page again.
So, I think the idea here is that I want to subclass the register view from django-registration. This is just where I'm not sure how to proceed. Is this the right direction? Should I test the user's authentication status here? Tips and advice welcomed!
Edit
I think this is the right track here: Django: Redirect logged in users from login page
Edit 2
Solution:
Create another app, for example, custom_registration, and write a view like this (mine uses a custom form as well):
from registration.views import register
from custom_registration.forms import EduRegistrationForm
def register_test(request, success_url=None,
form_class=EduRegistrationForm, profile_callback=None,
template_name='registration/registration_form.html',
extra_context=None):
if request.user.is_authenticated():
return HttpResponseRedirect('/')
else:
return register(request, success_url, form_class, profile_callback, template_name, extra_context)
I had to use the same function parameters, but otherwise just include the test, and if we pass it, continue to the main function.
Don't forget to put this in your URLConf either (again, this includes some stuff about my custom form as well):
top-level URLConf
(r'^accounts/', include('custom_registration.urls')),
(r'^accounts/', include('registration.urls')),
custom_registration.views
from django.conf.urls.defaults import *
from custom_registration.views import register_test
from custom_registration.forms import EduRegistrationForm
urlpatterns = patterns('',
url(r'^register/$', register_test, {'form_class': EduRegistrationForm}, name='registration.views.register'),
)

As far as I remember django-registration is using function-based views, so you can not really subclass them. The approach I usually follow is "overwriting" the original views (without modifying the django-registration app of course). This works like this:
Create another app (you could call it custom_registration or whatever you want)
This app need to contain another urls.py and in your case another views.py
Copy the original register view code to your new views.py and modify it, add a pattern to your urls.py to point to this view (use the same url pattern as in django-registration for this view)
Put an include to your projects urls.py of your new app urls.py before your are including the original django-registration app. This could look like this for example:
urlpatterns = patterns('',
...
url(r'^accounts/', include('custom_registration.urls')),
url(r'^accounts/', include('registration.backends.default.urls')),
...
)
This simply works since the first matching url pattern for /accounts/register will point to your new app, so it will never try to call the one from the original app.