I'm building a django-based media server where each person has a private account.
The person can upload media files such as mp3s. The media files will be uploaded to a directory '/media/ uploads/'+<username>+'/' with name of filename.mp3 I started working with this posting: http://groups.google.com/group/django-users/browse_thread/thread/6876b45961c6ade6/054fb073921ebd31?lnk=raot&pli=1 But it solves a slightly different challenge (posting above web root). It seems to require setting the path at 'compile' rather than run time. (I know python is interpreted.) I found a code snipped posted a few days ago that appears to address the issue, but I lack a deep enough understanding to know the one-line change I need to make to django-transfers based on this snippet. I'm guessing it is here; if request.user: self.get_backend = lambda: backend(UPLOAD_DIR=request.username) Please direct me to any resources that can further me along this journey or suggest how to solve my challenge. Thank you in advance, Jonathan =========================================================================================== import json from django.http import HttpResponse, HttpResponseBadRequest, Http404 from ajaxuploader.views import AjaxFileUploader from ajaxuploader.backends.local import LocalUploadBackend class MyAjaxFileUploader(AjaxFileUploader): def _ajax_upload(self, request): # check to see if the user info is passed print request if request.user: self.get_backend = lambda: backend(UPLOAD_DIR=request.username) if request.method == "POST": if request.is_ajax(): # the file is stored raw in the request upload = request is_raw = True # AJAX Upload will pass the filename in the querystring if it # is the "advanced" ajax upload try: filename = request.GET['qqfile'] except KeyError: return HttpResponseBadRequest("AJAX request not valid") # not an ajax upload, so it was the "basic" iframe version with # submission via form else: is_raw = False if len(request.FILES) == 1: # FILES is a dictionary in Django but Ajax Upload gives # the uploaded file an ID based on a random number, so it # cannot be guessed here in the code. Rather than editing # Ajax Upload to pass the ID in the querystring, observe # that each upload is a separate request, so FILES should # only have one entry. Thus, we can just grab the first # (and only) value in the dict. upload = request.FILES.values()[0] else: raise Http404("Bad Upload") filename = upload.name backend = self.get_backend() # custom filename handler filename = (backend.update_filename(request, filename) or filename) # save the file backend.setup(filename) success = backend.upload(upload, filename, is_raw) # callback extra_context = backend.upload_complete(request, filename) # let Ajax Upload know whether we saved it or not ret_json = {'success': success, 'filename': filename} if extra_context is not None: ret_json.update(extra_context) return HttpResponse(json.dumps(ret_json)) -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
