How can I view my HTML page in Python using Django

Question

I'm new into creating projects in Python using Django.

I'm creating a website in Python using the Django framework. I've created an "index" function in the "views.py" file to render the "index.html" file present in the "templates" folder. Below is the code for the "views.py" file.

from django.http import HttpResponse
from django.shortcuts import render


# Create your views here.
def index(request):
    return render(request,"index.html",{})

I've also added the navigation for the "index" page. Below is the code for "urls.py" file.

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

from gallery import views

urlpatterns = [
    path('admin/', admin.site.urls),
    path('index/', views.index,name="index")
]

Below is the code of templates section in settings.py file.

TEMPLATES = [
    {
        'BACKEND': 'django.template.backends.django.DjangoTemplates',
        'DIRS': [os.path.join(BASE_DIR,'templates'),], #path to templates folder in local system.
        'APP_DIRS': True,
        'OPTIONS': {
            'context_processors': [
                'django.template.context_processors.debug',
                'django.template.context_processors.request',
                'django.contrib.auth.context_processors.auth',
                'django.contrib.messages.context_processors.messages',
            ],
        },
    },
]

But when I try to open the "index" page URL, I'm not able to view the page.

Error screenshot for reference

Folder structure for reference:-

Gallery_Website
|--gallery
  |_ __init__.py
  |_ admin.py
  |_ apps.py
  |_ models.py
  |_ tests.py
  |_ views.py
|--Gallery_Website
  |_ __init__.py
  |_ settings.py
  |_ urls.py
  |_ wsgi.py
|--templates
  |_ index.html
|--db.sqlite3
|--manage.py

What could be wrong while defining the navigation for the "index" page?

Let me know if any other information is required.

Did you try to execute ' return render(request = request, template_name = 'index.html'' ' ? Some times, atleast in some of the versions you need to specify the parameter name in order to get the result in render() function Also, did you make sure that the index.html file is in the same directory as the views.py file ? — JimShapedCoding
– JimShapedCoding, Commented Nov 10, 2019 at 8:16
Please Clarify what are you getting on that page a ''error'' or something like that? — Dev
– Dev, Commented Nov 10, 2019 at 8:41
what is content of your index.html file and make sure that you have created templates folder in your gallery app and inside you have index.html page. — Subham
– Subham, Commented Nov 10, 2019 at 8:44

Mukul Kumar · Accepted Answer · 2019-11-10 09:28:27Z

1

In your settings.py file, add a value to you template dictionary key -- DIRS=[]

TEMPLATES = [
    {
        'BACKEND': 'django.template.backends.django.DjangoTemplates',
        'DIRS': [os.path.join(BASE_DIR, 'templates'),],
        'APP_DIRS': True,
        'OPTIONS': {
            'context_processors': [
                'django.template.context_processors.debug',
                'django.template.context_processors.request',
                'django.contrib.auth.context_processors.auth',
                'django.contrib.messages.context_processors.messages',
            ],
        },
    },
]

then create a folder "templates" where "manage.py" file resides. And in that templates folder, create a file "index.html". Then you will be a able to see your html page.

answered Nov 10, 2019 at 9:28

Mukul Kumar

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4 Comments

BBRK Over a year ago

Hi kumarm5, I've edited the post for your reference above.

Mukul Kumar Over a year ago

It seems like your requested url is : - '127.0.0.1:8000'. So just need to change the requested url to :- 127.0.0.1:8000/index (url ends with slash)

BBRK Over a year ago

Is there any way if the page can be viewed when I hit '127.0.0.1:8000' url ?

Mukul Kumar Over a year ago

Yes, you can import re_path and do that like this:- from django.urls import path, re_path urlpatterns = [ path('admin/', admin.site.urls), re_path('^', views.index,name="index") ]

Maharshi · Accepted Answer · 2019-11-10 10:30:14Z

1

You have to remove 'index/' in urls.py

Correct Code:

urlpatterns = [
     path('admin/', admin.site.urls),
     path('', views.index,name="index")
]

answered Nov 10, 2019 at 10:30

Maharshi

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