Passing Arguments to Django Views Without URL Parameters

Based on your question, you're trying to pass a return_path parameter to your view without it appearing in the URL. The {% url %} template tag is designed to generate URLs based on your URL patterns, so any parameters you pass will be included in the URL.

Solution: Use Query Parameters

The best approach for your scenario is to use query parameters instead of trying to pass the parameter directly through the URL tag:

DJANGOCopied
<a href="{% url 'jobs:change_contract_status' job.pk %}?return_path={{ request.path|urlencode }}">
    {{ job|contract_status_button }}
</a>

Then in your view, you can access this parameter:

PYTHONCopied
def change_contract_status(request, pk):
    # Your logic here
    
    # Get the return path from query parameters
    return_path = request.GET.get('return_path', '/')
    
    # After processing, redirect to the return path
    return redirect(return_path)

Alternative: Using the HTTP Referer

Another approach is to use the HTTP Referer header, which contains the URL of the page that linked to the current page:

PYTHONCopied
def change_contract_status(request, pk):
    # Your logic here
    
    # Get the referring URL or default to home
    return_path = request.META.get('HTTP_REFERER', '/')
    
    # After processing, redirect to the return path
    return redirect(return_path)

This approach doesn't require any changes to your template, but be aware that the Referer header might not always be available (users can disable it in their browsers).

URL Configuration

Your URL configuration remains the same:

PYTHONCopied
path('change_contract_status/<int:pk>/', v.change_contract_status, name='change_contract_status'),

The query parameters don't need to be defined in the URL pattern as they're automatically available in the request.GET dictionary.