I have such a urls.py configuration in project repository "forum"
# Project url
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r"^$", views.index, name="index"),
url(r'^article/', include('article.urls',namespace='article')),
]
and with article/urls.py as
#Article app's url config
urlpatterns = [
# show the article list
url(r"^list/(?P<block_id>\d+)$", views.article_list, name="article_list"),
I attempt to take "^article/list/1$" as home page instead of "views.index".
How to make it redirect to "^article/list/1$" when I issue request "127.0.0.1:8000"?
You can redirect from your views.index. In your views.py
from django.shortcuts import redirect
def index(request):
return redirect('article:article_list')
which should take you to article/list. You can include that block_id parameter in the redirect function.
try this
redirect(reverse('article:article_list', kwargs={'block_id':2}))
and make sure to add kwargs in function like this
article_list(request,**kwargs):
I assume you want to this to debug your article page.
You have two ways of doing this.
Either redirect your index view to article view
views.py
from django.shortcuts import redirect
def index(request):
#We force a redirect to article_list view, and we pass arguments.
#Notice it has to come first.
redirect ("article_list", article_list_argument_view=1)
#it will redirect to localhost:8000/article/list/1
'''
Your Home page code.
'''
Put the url directly before the include
urls.py
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r"^$", views.index, name="index"),
#The order is important.
url(r'^article/list/1$', "article_list", {'article_list_argument_view':1})
url(r'^article/', include('article.urls',namespace='article')),
]
Related
I would like to redirect all 404 pages to a home page. I try this but it don't work
app/views.py
from django.http import HttpResponse
from django.shortcuts import render, redirect
def home(request): return HttpResponse('<h1> HOME </h1>')
def redirectPNF(request, exception): return redirect('home')
app/urls.py
from . import views
urlpatterns = [ path('home', views.home, name="home"), ]
app/settings.py
handler404 = 'app.views.redirectPNF'
ALLOWED_HOSTS = ['127.0.0.1', 'localhost']
DEBUG = False
Just Add this line in urls.py instead of settings.py
Everything else seems ok.
It is also mentioned in the django documentation that setting handler variables from anywhere else will have no effect. It has to be set from URLconf
The default error views in Django should suffice for most web applications, but can easily be overridden if you need any custom behavior. Specify the handlers as seen below in your URLconf (setting them anywhere else will have no effect).
app/urls.py
from . import views
handler404 = 'app.views.redirectPNF' # Added this line in URLconf instead of settings.py
urlpatterns = [ path('home', views.home, name="home"), ]
I have a few functions in the view.py in my Django project:
Here is the views.py and urls.py under polls:
polls/views.py
from django.shortcuts import render
from django.http import HttpResponse
def index(request):
return HttpResponse("Hello, world. You're at the polls index.")
def search(request):
return HttpResponse("You're at the polls search.")
polls/urls.py
from django.urls import path
from . import views
from django.conf.urls import include, url
urlpatterns = [
path('', views.index, name='index'),
path('', views.search, name='search'),
]
I am able to get the index page, but have trouble to reach the page in the search function. I got the error below:
How do I access the search function in the views.py? Thanks!
edit your polls/urls.py as below:
urlpatterns = [
path('', views.index, name='index'),
path('search/', views.search, name='search'),
]
the first argument of the path is the url pattern.
I think you misunderstood the third argument(name). it has nothing to do with the url pattern, it's a name for the url, that'll be useful for url reversing. read the document for more information
Was hoping someone could point me in the right direction with this. I've tried nearly everything I can think of, but I can't seem to get it to work. I've got a set of URLs I'd like to match in Django:
www.something.com/django/tabs/
www.something.com/django/tabs/?id=1
Basically, I want to make it so that when you just visit www.something.com/django/tabs/ it takes you to a splash page where you can browse through stuff. When you visit the second URL however, it takes you to a specific page which you can browse to from the first URL. This page is rendered based on an object in the database, which is why the id number is there. I've tried to account for this in the URL regex, but nothing I try seems to work. They all just take me to the main page.
This is what I have in urls.py within the main site folder:
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^tabs/', include("tabs.urls")),
]
and within urls.py in the app's folder:
urlpatterns = [
url(r'\?id=\d+$', tab),
url(r'^$', alltabs)
]
Would anyone be so kind as to point me in the right direction? Thanks in advance!
You are not following the right approach here. Query paramers are used to change the behaviour of the page slightly. Like a added filter, search query etc.
What i would suggest is you have only one view and render different templates based on query parameters in the view.
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^tabs/', alltabs),
]
In your alltab views you can have something like this.
def alltabs(request):
if request.GET.get("id"):
id = request.GET.get("id")
your_object = MyModel.objects.get(id=id)
return render_to_response("tab.html", {"object":your_object})
return render_to_response("alltab.html")
Hope this helps
This is not the preferred 'django way' of defining urls patterns, I would say:-)
In the spirit of django would be something like
www.something.com/django/tabs/
www.something.com/django/tabs/1/
....
www.something.com/django/tabs/4/
and for this you define your url patterns within the app for example this way
tabs/urls.py:
from django.conf.urls import url
from . import views
urlpatterns = [
# ex: /tabs/
url(r'^$', views.index, name='index'),
# ex: /tabs/5/
url(r'^(?P<tab_id>[0-9]+)/$', views.detail, name='detail'),
# ex: /tabs/5/results/
url(r'^(?P<tab_id>[0-9]+)/results/$', views.results, name='results'),
]
and something similar in your views
tabs/views.py:
from django.shortcuts import get_object_or_404, render
from tabs.models import Tab
def index(request):
return render(request, 'tabs/index.html')
def detail(request, tab_id):
tab = get_object_or_404(Tab, pk=tab_id)
return render(request, 'tabs/detail.html', {'tab': tab})
...
You can follow this django tutorial for more details:
My urls.py is
from django.conf.urls import patterns,url
from rango import views
urlpatterns=patterns('',url(r'^$',views.index,name='index'))
urlpatterns=patterns('',url(r'^about/$',views.about,name='about'))
My views.py is
from django.shortcuts import render
from rango.models import Category
# Create your views here.
from django.http import HttpResponse
def index(request):
category_list = Category.objects.order_by('-likes')[:5]
context_dict={'categories':category_list}
return render(request, 'rango/index.html', context_dict)
def about(request):
return HttpResponse("go to index")
When I am trying to go to the address http://127.0.0.1:8000/rango I am getting page not found. But I am able to go to the address http://127.0.0.1:8000/rango/about.
When I remove the about url pattern in urls.py, I am able to go to the address http://127.0.0.1:8000/rango but not http://127.0.0.1:8000/rango/about, as the about url pattern does not exist.
I am unable to access both urls at once.
You have defined urlpatterns twice. The second patterns containing the about view replaces the first, which stops you accessing the index view.
Instead of,
urlpatterns=patterns('',url(r'^$',views.index,name='index'))
urlpatterns=patterns('',url(r'^about/$',views.about,name='about'))
it should be:
urlpatterns = patterns('',
url(r'^$', views.index, name='index'),
url(r'^about/$', views.about, name='about'),
)
In Django 1.7+, you don't need to use patterns any more, so you can simplify it to
urlpatterns = [
url(r'^$', views.index, name='index'),
url(r'^about/$', views.about, name='about'),
]
On my home page I have a link to a registration page.
In my browser, the home page is at 127.0.0.1:8000/app_name/
When I click the link to register, it takes me to 127.0.0.1:8000/app_name/app_name/register/, which gives a 404 error.
But it should take me to 127.0.0.1:8000/app_name/register/, which displays the proper html.
In my views.py, for the Registration page I have:
return render(request, 'app_name/register.html', {'form': form})
In my app urls.py, for the Registration page I have:
url(r'^register/$', views.register, name='register'),
Why am I getting the duplicate app_name in my url?
Update:
Here is my Project urls.py:
from django.conf.urls import patterns, include, url
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
url(r'^app_name/', include('app_name.urls')),
url(r'^admin/', include(admin.site.urls)),
)
Here is my app_name urls.py:
from django.conf.urls import patterns, url
from app_name import views
urlpatterns = patterns('',
url(r'^$', views.index, name='index'),
url(r'^register/$', views.register, name='register'),
)
Please add templates dir inside of your project dir and add this path in settings.py(you can add absolute and relate path) add your html pages in this templates folder
read my answer to add path in settings.py
and in your html for link to registration page
Sign Up
your view
return render(request, 'register.html', {'form': form})
Are you declaring the url in your app urls.py or the projects? (I'm wondering how many patterns declarations it's going through)