urls.py django 2.0.2 need parameters

Question

i'm learning django and i'm making a blog tutorial but it's from older version that i have, i have version 2.0.2 and i'm not understand documentation my problem is that i dont know how configure my urls.py this is my three proyect:

i need to put archive.html in 127.0.0.1:8000/ and this is my urls code

codigofacilito/blog.urls.py :

"""codigofacilito URL Configuration

The urlpatterns list routes URLs to views. For more information please see: https://docs.djangoproject.com/en/2.0/topics/http/urls/
Examples:
Function views
1. Add an import: from my_app import views
2. Add a URL to urlpatterns: path('', views.home, name='home')
Class-based views
1. Add an import: from other_app.views import Home
2. Add a URL to urlpatterns: path('', Home.as_view(), name='home')
Including another URLconf
1. Import the include() function: from django.urls import include, path
2. Add a URL to urlpatterns: path('blog/', include('blog.urls'))
"""

from django.contrib import admin<br/>
from django.urls import path<br/>
from . import views<br/>


urlpatterns = [
    path('admin/', admin.site.urls),

]

and codigofacilito/codigofacilito.urls.py:

from django.contrib import admin
from django.urls import path
from . import views



enter code here`urlpatterns = [
    path('admin/', admin.site.urls),

]

Can you be more specific.. What is that you want to do, and what is already done? I am myself a django starter and propably went though the same trouble as you are now — Vipin Mohan
If you don't know how to map a URL to a view, you should read the tutorial where all this is clearly explained. — Daniel Roseman
In Django 2.0.2 you can still use url() as your tutorial does. You don't have to switch to path(). — Alasdair

Rohit Chopra Rohit Chopra · Accepted Answer · 2018-03-05T12:26:00

from django.views.generic import TemplateView

urlpatterns = [
    path('', TemplateView.as_view(template_name = 'archive.html')),
    path('admin/', admin.site.urls),

]

put this in your project's urls.py, this will directly render the archive.html when you visit 127.0.0.1:8000/ in your browser.

but if you want to render data from backend in this html page then i suggest you to use views (function based , class based, etc..)

then you just have to import the views in url file and specify path for that.

from your_app.views import your_view

urlpatterns = [
    path('/', your_view),
]

urls.py django 2.0.2 need parameters

1 Answers