2016-04-03 78 views
7

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 
+1

您能否包含'serve'视图中的代码? – xthestreams

回答

21

您参数文件_根下划线错过。但在生产中使用serve是个好主意。使用类似这样的代码:

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

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 
+0

工作很好。谢谢 – jsm1th

+0

它对我来说很顺利。 thx @Sergey Gornostaev – gustav