django download csv file using a link

Question

I am a new to django and python. Need some guidance in this quest.

Case: When the user hits the submit button on a form, it should display Success page and a link wh

Answer 1

In your urls.py change

urls.py url(r'^static/(?P.*)$', send_file)

to

urls.py url(r'^static/example.xls$', send_file)

In the first one, you are also passing everything after the / to the view as another parameter, but your view does not accept this parameter. another option would be to accept this parameter in the view:

def send_file(request, path):
    ...

but since the path to your xls file is hard coded, I don't think you need that.