I am learning django and I am trying to add a new url to '/'
here is my urls.py:
from django.contrib import admin
from django.urls import path
from blog import views as blog_views
urlpatterns = [
path(r'^$', blog_views.index),
path('admin/', admin.site.urls),
]
and here is the index method from blog/views:
def index(request):
return HttpResponse("Hey There")
But when I go to '/' route, I get a 404 response. It seems that it fails at import from blog import views as blog_views in urls.py.
What is going wrong?
Here is my project structure:
Here is the error I get:
In Django 2.x there is a path function instead of django's 1.x url function
the path function doesn't accept Regular Expressions it only accepts normal text
So to make a url for the home page with path function you only need to write your url path like this :
urlpatterns = [
path('', blog_views.index), # http://localhost/
path('admin/', admin.site.urls),
]
read more about Django 2.x urls here:
https://docs.djangoproject.com/en/dev/ref/urls/
Related
I am following the django tutorial (found here):
views.py code:
from django.http import HttpResponse
def index(request):
return HttpResponse("Hello, world. You're at the poll index.")
urls.py code (this is inside polls app urls.py file):
from django.urls import path, include
from django.conf import settings
from . import views
urlpatterns = [path(r'^$', views.index, name='index'), ]
urls.py code ( this is root urls.py file code):
from django.contrib import admin
from django.urls import include, path
urlpatterns = [path('polls/', include('polls.urls')),path('admin/',admin.site.urls), ]
Here is my run command : python manage.py runserver 8080
I tried to run it today, but I am getting the following error:
Page not found (404)
Request Method: GET
Request URL: http://35527a91f40c4e228d6c464d8a8c8487.vfs.cloud9.eu-west-1.amazonaws.com/
Using the URLconf defined in PollApp.urls, Django tried these URL patterns, in this order:
poll/
admin/
The empty path didn't match any of these.
You're seeing this error because you have DEBUG = True in your Django settings file. Change that to False, and Django will display a standard 404 page.
You are using regex-syntax with path, you should use the empty string instead:
from . import views
urlpatterns = [
path('', views.index, name='index'),
]
Furthermore, you use a poll/ prefix here, so that means you either need to visit localhost:8000/poll/, or change this to an empty prefix:
urlpatterns = [
path('', include('polls.urls')),
path('admin/', admin.site.urls),
]
Djano errors are self-explanatory, follow its instructions, and are good to go.Also is there the Indentation in your first return ?
from . import views
urlpatterns = [
path('', views.index, name='index'),
]
I am a total beginner in "django" so I'm following some tutorials currently I' am watching https://youtu.be/JT80XhYJdBw Clever Programmer's tutorial which he follows django tutorial
Everything was cool until making a polls url
Code of views.py:
from django.shortcuts import render
from django.http import HttpResponse
def index(request):
HttpResponse("Hello World.You're at the polls index")
Code of polls\urls.py:
from django.urls import path
from . import views
urlpatterns = [
path('', views.index, name='index'),
]
Code of Mypr\urls.py:
from django.contrib import admin
from django.urls import path, include
urlpatterns = [
path('admin/',admin.site.urls),
path('polls/',include('polls.urls')),
]
I don't get it I did the same thing but I'm getting error not only polls.In one turtorial he decided to make blog,and again the same error:
Page not found (404)
Request Method: GET
Request URL: http://127.0.0.1:8000/polls/
Using the URLconf defined in Mypr.urls, Django tried these URL patterns, in
this order:
admin/
The current path, polls/, didn't match any of these.
Please my seniors help me.
Note:I'm using the latest versions of django,windows and as editor I'm using Pycharm.
Already tried(and did not work):
from django.urls import path
from polls.views import index
urlpatterns = [
path('', index, name='index'),
]
Page not found (404)
Request Method: GET
Request URL: http://127.0.0.1:8000/polls/
Using the URLconf defined in Mypr.urls, Django tried these URL patterns, in this order:
admin/
The current path, polls/, didn't match any of these.
You're seeing this error because you have DEBUG = True in your Django settings file.
Change that to False, and Django will display a standard 404 page.
try to access polls/ URL in your browser then you will access the page
Because you have accessed the URL of your project, you have to go to this URL to access your app
try by changing your code like this
from django.urls import path
from polls.views import index
urlpatterns = [
path('', index, name='index'),
]
I was facing the same error. Feel kinda dumb after figuring out how to solve this.
Your urls.py content is all right
from django.contrib import admin
from django.urls import include, path
urlpatterns = [
path('polls/', include('polls.urls')),
path('admin/', admin.site.urls),
]
But it must be included in the mysite\mysite\urls.py and not in mysite\urls.py.
That means the inner mysite/ directory is the actual Python package for your project. Its name is the Python package name you’ll need to use to import anything inside it.
I am doing django course from udemy, i did one experiment. Below is my folder structure
Project
appTwo
urls.py
ProTwo
urls.py
appTwo/urls.py
from django.conf.urls import url
from appTwo import views
urlpatterns = [
url(r'^$',views.help,name='help'),
url(r'^$',views.users,name='users'),
]
ProTwo/urls.py
from django.contrib import admin
from django.urls import path
from django.conf.urls import url,include
from appTwo import views
urlpatterns = [
url(r'^$',views.index,name='index'),
url(r'^help/',include('appTwo.urls')),
url(r'^users/',include('appTwo.urls')),
path('admin/', admin.site.urls),
]
Now when i try to open the page users by http://127.0.0.1:8000/users it opens the page help.html. For http://127.0.0.1:8000/help it opens help page. When I comment the first entry in urlpatterns in urls.py it opens the users page even if i try to open help page. Can anyone please guide me what wrong I am doing or its working as expected.
you need to use different patterns for each View:
urlpatterns = [
url(r'^help$', views.help, name='help'),
url(r'^users$', views.users, name='users'),
]
the name attribute is only useful for the concept of reversing, maybe further in your course?
also, the r'' strings in python are regular expressions, you might want to learn more about them.
I got the answer for this. Below two files need to be changed
appTwo/urls.py
from django.conf.urls import url
from appTwo import views
urlpatterns = [
url(r'^help$',views.help,name='help'),
url(r'^users$',views.users,name='users'),
]
ProTwo/urls.py
from django.contrib import admin
from django.urls import path
from django.conf.urls import url,include
from appTwo import views
urlpatterns = [
url(r'^$',views.index,name='index'),
url(r'^appTwo/',include('appTwo.urls')),
path('admin/', admin.site.urls),
]
Explanation:
When I type in browser "basic url" i mean the address here it is http://127.0.0.1:8000/, it will go to ProTwo/urls.py which is the project folder. This will open the index page as per line url(r'^$',views.index,name='index'). If you need two configure two different page, give url(r'^appTwo/',include('appTwo.urls')), in the ProTwo/urls.py. This will call appTwo/urls.py. Now for help type http://127.0.0.1:8000/appTwo/help and for users type http://127.0.0.1:8000/appTwo/users.
after wasting saturdays eve googling for a solution to finally make a Django server work, i need your assistance..
I first of all want to set up my project in that way that http://127.0.0.1:8000/ redirects me to the index.html site. But somehow I am not able to run the Django server within my virtualenv (access denied).
I handled error over error in the past few hours (inserted a Secret key, inserted silenced_system_checks since E408/09/10 occured as errors before the current error) and here I stuck now. I am not capable to understand the prompt error at all. I assume that Django wants to start the server but can't find a file/html to return?
urls.py // dassocc_app dir
from django.urls import path
from django.urls import include
from django.conf.urls import url
from . import views
urlpatterns = [
path('/dasocc_site/dasocc_app/templates/', include("dasocc_app.views")),
path('', views.liga, name="index"),
]
views.py
import requests
from django.shortcuts import render
def liga(request):
liga = ['1. Bundesliga', 'Premier League', 'La liga']
return render(request, 'dasocc_app/templates/index.html', {'liga': liga})
urls.py // dasocc_site dir
from django.urls import path
from django.urls import include
from dassoc_app import views
urlpatterns = [
url(r'^$', views.liga, name='index')
]
enter image description here
Your troublemaker is the line path('', views.index, name="index"). It cannot find a function called index in your views.py.
Assuming the function you want to call is liga() you will have to write
path('', views.liga, name="index").
Or you could rename your liga function to index
#2
Kindly change your dassoc_site.urls:
dassoc_site/urls.py
from dasocc_app import views
from django.conf.urls import url, include
urlpatterns = [
url(r'^$', views.liga, name='index'),
url(r'^dassoc-app/', include('dassoc_app.urls')),
]
dassoc-app/urls.py
from django.conf.urls import url
from dasocc_app import views
urlpatterns = [
# Where home is some random view from your dassocc-app
url(r'^$', views.home, name='home')
]
I have a problem with Django. Actually I want to create a login panel in my new site but when I try to write address the debugger respond me a error:
Page not found (404)
Request Method: GET
Request URL: http://localhost:8000/account/login
Using the URLconf defined in firmowa.urls, Django tried these URL patterns, in this order:
admin/
^account/
The current path, account/login, didn't match any of these.
My firmowa.urls is
from django.contrib import admin
from django.urls import path, include
urlpatterns = [
path('admin/', admin.site.urls),
path(r'^account/', include('account.urls')),
]
And the project urls.py is:
from django.urls import path
from . import views
urlpatterns = [
path(r'^login/$', views.user_login, name='login'),
]
I'm looking a solution from few hours but still nothing.
path is a new feature in Django 2.0 and works different than the old url. Don't use regex in path. If you need regex, use re_path instead, but in your case there's no need for that.
Change your code to:
urlpatterns = [
path('login/', views.user_login, name='login'),
]
etc. and it should work.
More on the topic here and here.
from django.urls import path
from . import views
from django.conf.urls import url
urlpatterns = [
url( r'^$', views.index , name='index') ,
]