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rakesh kumar
rakesh kumar

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How to implement 403 forbidden error in django

Sure! Here's an example of how you can implement a 403 Forbidden error handling in Django:

Step 1: Define Forbidden Error View
Create a file called in your Django app and define a view to handle the 403 Forbidden error. This view can be a simple function-based view or a class-based view. Here's an example of a function-based view:

from django.shortcuts import render

def forbidden_error(request, exception):
    return render(request, '403.html', status=403)
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Step 2: Define URLs
In your app's file, add a URL pattern for the forbidden error view:

from django.urls import path
from .views import forbidden_error

urlpatterns = [
    # Your other URL patterns

    # URL pattern for handling 403 Forbidden error
    path('forbidden/', forbidden_error, name='forbidden'),

handler403 = 'yourapp.views.forbidden_error'
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Step 3: Create 403 Template
Create a template called 403.html in your app's template directory. This template will be rendered when a user encounters a 403 Forbidden error. Customize the template to suit your needs.

Step 4: Update Middleware Settings
In your project's file, make sure the django.middleware.common.CommonMiddleware middleware class is enabled. This middleware is responsible for handling the 403 Forbidden error.

    # Other middleware classes
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Step 5: Trigger a 403 Forbidden Error
To test your implementation, you can manually trigger a 403 Forbidden error by raising the PermissionDenied exception in your views or using the permission_denied method. For example:

from django.core.exceptions import PermissionDenied

def my_view(request):
    if not has_permission(request):
        raise PermissionDenied
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# Rest of your view logic
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That's it! With the above steps, you have implemented handling of the 403 Forbidden error in Django. Users who encounter a forbidden error will be redirected to the forbidden_error view, which will render the 403.html template with the appropriate HTTP status code.

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