我是使用Django的新手,我正在尝试开发一个网站,用户可以在其中上传许多excel文件,然后将这些文件存储在媒体文件夹Webproject / project / media中。
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)
然后,该文档会与它们上载的任何其他文档一起显示在列表中,你可以单击这些文档,它会显示有关它们的基本信息以及他们上载的excelfile的名称。从这里,我希望能够使用链接再次下载相同的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)
但我得到了错误,serve()得到了一个意外的关键字参数’document root’。谁能解释如何解决此问题?
要么
解释如何获取上传的文件以进行选择和使用
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
你错过了参数文档_ root 下划线。但是serve在生产中使用它是个坏主意。使用类似这样的东西:
serve
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