Develop Reference

i am trying to learn django from django documentation but i am getting an error message in the first part of the documentation

Using the URLconf defined in blogs.urls, Django tried these URL patterns, in this order:
admin/
The current path, polls/, didn’t match any of these.
this is the error i am getting after i wrote the following code:
for urls.py
from django.contrib import admin
from django.urls import include, path
urlpatterns = [
path('polls/', include('polls.urls')),
path('admin/', admin.site.urls),
]
for polls/urls.py
from django.urls import path
from . import views
urlpatterns = [
path('', views.index, name='index'),
]
3.for polls/views.py
"""
from django.http import HttpResponse
def index(request):
return HttpResponse("Hello, world. You're at the polls index.")
"""
how i can i remove this error in vs code

Django page not found, help would be appreciated

I was taking a youtube tutorial on making a basic website using django and I got this error when coding:
Page not found (404)
Request Method: GET
Request URL: http://127.0.0.1:8000/
Using the URLconf defined in mysite.urls, Django tried these URL patterns, in this order:
app/
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.
This is my code:
For urls.py/mysite:
from django.contrib import admin
from django.urls import path, include
urlpatterns = [
path('app/', include('myapp.urls')),
path('admin/', admin.site.urls),
]
For views.py:
from django.shortcuts import render
from django.http import HttpResponse
def index(request):
return HttpResponse("Hello, world!")
For urls.py/myapp:
from django.urls import path
from . import views
urlpatterns = [
path('', views.index, name='index'),
]

You need to add root level path to access the path you specified http://127.0.0.1:8000/:
urls.py/mysite
urlpatterns = [
path('', views.index, name='home'),
path('app/', include('myapp.urls')),
path('admin/', admin.site.urls),
]
views.py/mysite
def index(request):
return HttpResponse('This is home page')

Since you specified:
path('app/', include('myapp.urls')),
It means all the paths in myapp.urls are prefixed with app/. So you can access the index view with: http://127.0.0.1:8000/app/. Or if you want to access the index view with http://127.0.0.1:8000/ you rewrite the path to:
path('', include('myapp.urls')),

Routing with Django and React

I'm very new to working with Django and I've been relying on some tutorials to link this one to React, but the problem is that initially (when I open 127.0.0.1:8000) React loads the routes perfectly, then when I reload the page Django tries to interpret the path from urls.py and obviously can't find it.
The error is:
Page not found (404) Using the URLconf defined in memberstack.urls, Django tried these URL patterns, in this order:
admin/
api/token-auth/
core/
I hope you can help me, thanks in advance
my_project/urls.py
from django.contrib import admin
from django.urls import path, include
from rest_framework_jwt.views import obtain_jwt_token
from frontend.views import index
urlpatterns = [
path('', include('frontend.urls')),
path('admin/', admin.site.urls),
path('api/token-auth/', obtain_jwt_token),
path('core/', include('core.urls')),
]
frontend/urls.py
from django.urls import path
from . import views
urlpatterns = [
path('', views.index),
]
frontend/views.py
from django.shortcuts import render
def index(request):
return render(request, 'frontend/index.html')

For this, you'll have to use a catch-all in order for React to handle all the routing instead of django there.
from django.urls import path, re_path
from . import views
urlpatterns = [
path('', include('frontend.urls')),
path('admin/', admin.site.urls),
path('api/token-auth/', obtain_jwt_token),
path('core/', include('core.urls')),
re_path(r'^(?:.*)/?$', include('frontend.urls')),
]
Or
urlpatterns = [
path('', views.index),
re_path(r'^(?:.*)/?$', views.index)
]
I think the better practice would be to implement Django-Rest-Framework and build them separately.

Page not found (404) Request Method: GET Request URL: http://127.0.0.1:8000/catalog/

I got the follwoing error whenever I run the project and automatically catalog/ is added to the url:
Page not found (404)
Request Method: GET
Request URL: http://127.0.0.1:8000/catalog/
Using the URLconf defined in first_project.urls, Django tried these URL patterns, in this order:
[name='index']
admin/
first_app/
The current path, catalog/, 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.
Here is my project urls.py
from django.contrib import admin
from django.urls import path, include
from first_app import views
urlpatterns = [
path('', views.index, name='index'),
path('admin/', admin.site.urls),
path('first_app/', include('first_app.urls')),
]
Here is first_app urls.py code:
from django.urls import path
from first_app import views
urlpatterns = [
path('', views.index, name='index'),
]
How to get the index page as a default and get rid of the catalog.
Here is views.py file:
from django.shortcuts import render
from django.http import HttpResponse
from first_app.models import Topic, AccessRecord, Webpage
# Create your views here.
def index(request):
webpages_list = AccessRecord.objects.order_by('date')
date_dict = {'access_records': webpages_list}
return render(request, 'first_app/index.html', context=date_dict)

Hey Mak in your code u didnt mention any url related to /catalog/ so you are getting that 404 error message. Error 404 means that the page couldnt found by the server add /catalog/ in your url patterns so that the server could know which page is to be shown if /catalog/ is requested . And as far as ur proble"and automatically catalog/ is added to the url:" may be its your browser autocompleting the address field so once plesase check it. if the request is only /catalog then ur myproject urls.py file should be something like this
from django.contrib import admin
from django.urls import path, include
from first_app import views
urlpatterns = [
path('', views.index, name='index'),
path('admin/', admin.site.urls),
path('first_app/', include('first_app.urls')),
path('catalog',views.your_function_name)
]
if the request is first_app/catalog then ur first_app urls.py file should be something like this in url patterns
urlpatterns = [
path('/catalog', views.catalog, name='catalog'),
]

from django.contrib import admin
from django.urls import path, include
from first_app import views
urlpatterns = [
path('admin/', admin.site.urls),
path('first_app/', include('first_app.urls')),]
and
from django.urls import path
from first_app import views
urlpatterns = [
path('', views.index, name='index'),
]
>>>>> if your templates files :
templates/index.html
def index(request):
webpages_list = AccessRecord.objects.order_by('date')
date_dict = {'access_records': webpages_list}
return render(request, 'index.html', context=date_dict)

Page not Found (404) in django

Am using django I got this error Page not Found (404) Request Method: GET Request URL: http://127.0.0.1:8000/ Using the URLconf defined in VibezT.urls, Django tried these URL patterns, in this order
music/
admin/
The empty path didn't match any of these.
VibezT\urls.py
from django.contrib import admin
from django.urls import include, path
urlpatterns =[
path('music/', include('music.urls')),
path('admin/', admin.site.urls),
]
For music URLs I have this code
music\urls.py
from django.urls import path
from . import views
urlpattern = [
path(r'^$',/ views.index, name='index')
]
For views i have this code
from django.http import HttpResponse
def index(request):
return HttpResponse("Hello World")

You have not defined a URL for / in your code. Your code is defined for music/
You need to go to http://127.0.0.1:8000/music/ to find your view.
Also, I do not know why you are structuring your URL like this if you are using path. path(r'^$',/ views.index, name='index')
Just replace that with this:
path('', views.index, name='index'),
in your music\urls.py