我对使用Django很陌生,我试图开发一个用户可以上传一些excel文件的网站,然后这些文件存储在一个媒体文件夹中Webproject/project /媒体。Django下载文件
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)
该文件,然后显示在列表中与他们上传的任何其他文件,您可以点击进入沿,它会显示有关这些基本信息,他们已经上传的excelfile的名称。在这里,我希望能够下载同一再次使用链接excel文件:
<a href="/project/download"> Download Document </a>
我的网址是
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)
,但我得到的错误,服务()得到了一个意想不到的关键字参数“文档根目录'。谁能解释如何解决这个问题?
OR
解释我怎样才能到要选择的上传文件,并使用服
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
您能否包含'serve'视图中的代码? – xthestreams