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 the comments Ofri Raviv. you mentioned that its giving you a

TypeError: an integer

which is because while creating FileWrapper u are passing two parameters out of which the second one[optional] is supposed to be integer but u passed 'rb'

wrapper = FileWrapper(file(filename),"rb")

Which should actually be written as ('rb' is the parameter to File)

wrapper = FileWrapper(file(filename,"rb"))

So it was just a misalignment of braces, but makes it hard to debug sometimes.

Answer 2

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.

Answer 3

Now it works but i had to change file extension from excel (.xls) to csv.

My urls.py=url(r'^static/example.txt', send_file)
My HTML link=<a href="../static/example.txt">Download CSV File</a>
My view.py

def send_file(request):

  import os, tempfile, zipfile
  from wsgiref.util import FileWrapper
  from django.conf import settings
  import mimetypes

  filename     = "C:\ex2.csv" # Select your file here.
  download_name ="example.csv"
  wrapper      = FileWrapper(open(filename))
  content_type = mimetypes.guess_type(filename)[0]
  response     = HttpResponse(wrapper,content_type=content_type)
  response['Content-Length']      = os.path.getsize(filename)    
  response['Content-Disposition'] = "attachment; filename=%s"%download_name
  return response