["python","django"]
Django download a file

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(): form.save() return render_to_response('project/upload_successful.html') else: form = FileForm() args = {} args.update(csrf(request)) 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'^$', ListView.as_view(queryset=Post.objects.all().order_by("-date")[:25], template_name="project/project.html")), url(r'^(?P<pk>\d+)$', 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? OR 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(fh.read(), content_type="application/vnd.ms-excel") response['Content-Disposition'] = 'inline; filename=' + os.path.basename(file_path) return response raise Http404