I've installed django version 2.0 and default for urls is path I know that if in django 1.x can be urls and follow with regular expression, how do I use path for parameters instead of urls, and so I dont want to use urls because django by default add path not urls if its can be or use urls still I want to know how to use a path with parameters in django 2.0
here's my code
from django.urls import include, path
from . import views
urlpatterns = [
path('', views.articles_list),
path('add', views.articles_add),
path('edit', views.articles_edit)
]
path('edit/<int:id>', views.articles_edit)
you can add parameters like this
in view
def edit(request, id):
Application urls.py File Add Following code
path('blog/< int:blogid >', views.blog)
Application views.py file, add following code:
def blog(request, blogid):
Make sure both place name should be same (blogid)
Related
I'm a newbie to Django and I know this probably has been asked alot of times.
So basically what's happening is when I try to create a new project and whenever I'm trying to run my server, by default it's opening http://127.0.0.1:8000/catalog/ and not http://127.0.0.1:8000/.
Even if I run the server with my other projects, I'm facing the same error.
I followed this django basics tutorial on https://developer.mozilla.org/en-US/docs/Learn/Server-side/Django/skeleton_website
Idk but somehow I think it's default address is set to http://127.0.0.1:8000/catalog/.
Here's the link to the repo for the project:
https://github.com/Fanceh/django-404-error
Here's my project's urls.py:
from django.contrib import admin
from django.urls import path, include
from testuapp import urls
urlpatterns = [
path('admin/', admin.site.urls),
path('',include("testuapp.urls"))
]
Here's the code in my testuapp urls.py:
from django.urls import path
from . import views
urlpatterns = [
path('', views.testu),
]
Here's my webapp's views.py file:
from django.shortcuts import render
# Create your views here.
def testu(request):
render(request, 'Greetings!')
Is there any way I can change it?
Regards
From the tutorial link it mentions the redirect.
Any request for the root URL, will redirect you to /catalog.
Screenshot from the tutorial below.
HTH
so i assume you are below url pattern structure in your testuapp project.
urlpatterns = [
path('catalog',include("views.catalog"))
]
views.calalog is the name of the method in your view file.
Ok I think I figured it out, it's just the chrome cache. I cleared it and bam it's working!
I'm following a django tutorial that is a little outdated and in the urls.py file of the first app's directory we need to configure where to direct django to for any url starting with 'notes/'.
There are two separate 'apps' inside the project. I'm in the first one, not notes.
This is the code currently. I added include to import statement:
from django.contrib import admin
from django.urls import path, include
urlpatterns = [
path('admin/', admin.site.urls),
path(url(r’^notes/‘, include('notes.urls'))),
]
Inside the urlpatterns object, on the first line, path('admin/', admin.site.urls), comes predefined, but I need to add a redirect such that django goes to a different app, called 'notes' and searches there for its entry point, since all of the urls will begin with ‘notes/’.
The tutorial says to use a regular expression and uses this code:
url(r’^notes/‘, include(notes.urls))
so that any url that starts with 'notes/' should be redirected to this other file notes.urls.
However the predefined ones that currently come out of the box with a django project start with path.
I enclosed my notes/n redirect line in path, but not sure if this is correct. Should I instead directly write:
url(r’^notes/‘, include(notes.urls))
Also, do I need to delete the first line provided?
path('admin/', admin.site.urls),
The tutorial has:
urlpatterns = patterns('',
url(r’^notes/‘, include(notes.urls)),
)
and no admin urls line. It's from 2014 I believe.
Simply do:
path('notes/', include('notes.urls'))
I want to write a reusable Django application.
I tell my users to add the following to their urls.py
path('slack/', include(('slack_integration.urls', 'slack_integration'), namespace='slack_integration'),
And in my urls.py I want to have a view login_callback.
Now in my view, I need to get a value of slack_integration:login_callback.
I can trust the user that he/she will integrate it with slack_integration prefix and use it. But is this the best practise? Can I somehow get the name of the namespace for the app if user chooses a different name for it?
Thanks a lot!
Using namespace= within urls.py files is no longer supported, as it moves something specific to the Django app outside of the Python package that is the Django app.
The best practice now is to define the app_name within the urls.py file inside the Django app.
The old way: DON'T DO THIS (pre-Django 2.0)
the root urls.py
path('slack/', include(('slack_integration.urls', 'slack_integration'), namespace='slack_integration'),
The new way: DO THIS! (Django 2.0+)
the root urls.py
from django.urls import path, include
urlpatterns = [
path('slack/', include(('slack_integration.urls', 'slack_integration')),
]
slack_integration/urls.py
from django.urls import path
app_name = "slack_integrations"
urlpatterns = [
path('', HomeView.as_view(), name='home'),
]
As you can see, this keeps the namespace for the patterns within the app itself, along with the templates most likely to use it. The days of extra instructions on how to include an app are over! Good luck.
Hello StackOverflow community,
I'm currently learning how to use the library Django combined with Python. However, I've ran into some issues which are somehow strange. My situation is the following. I have a project called "animals" which is the base of the Django application. My app is called "polls". Then I've defined the following views.
polls/views.py
from django.shortcuts import render
from django.http import HttpResponse
def animals(request, animal_id):
return HttpResponse("%s" % animal_id)
def index(request):
return HttpResponse('Hello world')
So far, so good. Unfortunatly I can't say the same about the urlpatterns.
polls/urls.py
from django.urls import path
from . import views
urlpatterns = [
path('',views.index,name='index'),
# Redirects to localhost:8000/polls/<animal_id>
path('<int:animal_id>/',views.animals,name='animals')
]
Whenever I want to navigate to the url "localhost:8000/polls/10/" Django reminds me of the fact, that the url is not accepted by the application and a 404 Error is thrown inside my browser. Am I missing something here?
UPDATE
I've managed to resolve the problem by fixing a rather trivial error. While the polls/urls.py was alright, the problem lay inside the animals/urls.py file. This file looked like this:
animals/urls.py
from django.conf.urls import url
from django.contrib import admin
from django.urls import include,path
urlpatterns = [
url(r'^admin/', admin.site.urls),
path('polls',include('polls.urls')),
]
As one can see, the "polls" path is not finished with a "/" sign. This indicates to Django that my desired route would be localhost:8000/polls where is any integer I add to the url. However, you have to add the slash at the end. Otherwise Django won't work as expected.
I am trying to create an index page for my home app using Django. I would like for the urls.py file to point directly to an index.html file in my templates folder within the home app. Therefore, I have the following:
urlpatterns = patterns('',url(r'^$',template_name='index.html'),)
This gives me an error when I load the page, however.
url() got an unexpected keyword argument 'template_name'.
All the resources I have looked at have a view instead of directly linking to an html, but I simply want to go to an index page. How can I achieve this? Thank you!
As per Django documentation:
This would be a proper way of doing it:
from django.conf.urls import patterns
from django.views.generic import TemplateView
urlpatterns = patterns('',
(r'^about/', TemplateView.as_view(template_name="about.html")),
)