Django download a file

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I’m quite new to using Django and I am trying to develop a website where the user is able to upload a number of excel files, these files are then stored in a media folder Webproject/project/media.

def upload(request):
    if request.POST:
        form = FileForm(request.POST, request.FILES)
        if form.is_valid():
            return render_to_response('project/upload_successful.html')
        form = FileForm()
    args = {}
    args['form'] = form

    return render_to_response('project/create.html', args)

The document is then displayed in a list along with any other document they have uploaded, which you can click into and it will displays basic info about them and the name of the excelfile they have uploaded. From here I want to be able to download the same excel file again using the link:

 <a  href="/project/download"> Download Document </a>

My urls are

 urlpatterns = [

              url(r'^$', ListView.as_view(queryset=Post.objects.all().order_by("-date")[:25],
              url(r'^(?P<pk>d+)$', DetailView.as_view(model=Post, template_name="project/post.html")),
              url(r'^upload/$', upload),
              url(r'^download/(?P<path>.*)$', serve, {'document root': settings.MEDIA_ROOT}),

          ] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

but I get the error, serve() got an unexpected keyword argument ‘document root’. can anyone explain how to fix this?


Explain how I can get the uploaded files to to be selected and served using

def download(request):
    file_name = #get the filename of desired excel file
    path_to_file = #get the path of desired excel file
    response = HttpResponse(mimetype='application/force-download')
    response['Content-Disposition'] = 'attachment; filename=%s' % smart_str(file_name)
    response['X-Sendfile'] = smart_str(path_to_file)
    return response


You missed underscore in argument document_root. But it’s bad idea to use serve in production. Use something like this instead:

import os
from django.conf import settings
from django.http import HttpResponse, Http404

def download(request, path):
    file_path = os.path.join(settings.MEDIA_ROOT, path)
    if os.path.exists(file_path):
        with open(file_path, 'rb') as fh:
            response = HttpResponse(, content_type="application/")
            response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path)
            return response
    raise Http404

Source: stackoverflow