serving files using django - is this a security vulnerability - django

Question

I'm using the following code to serve uploaded files from a login secured view in a django app.
Do you think that there is a security vulnerability in this code? I'm a bit concerned about that the user could place arbitrary strings in the url after the upload/ and this is directly mapped to the local filesystem.
Actually I don't think that it is a vulnerability issue, since the access to the filesystem is restricted to the files in the folder defined with the UPLOAD_LOCATION setting.
UPLOAD_LOCATION = is set to a not publicly available folder on the webserver
url(r'^upload/(?P<file_url>[/,.,\s,_,\-,\w]+)', 'project_name.views.serve_upload_files', name='project_detail'),
#login_required
def serve_upload_files(request, file_url):
import os.path
import mimetypes
mimetypes.init()
try:
file_path = settings.UPLOAD_LOCATION + '/' + file_url
fsock = open(file_path,"r")
file_name = os.path.basename(file_path)
file_size = os.path.getsize(file_path)
print "file size is: " + str(file_size)
mime_type_guess = mimetypes.guess_type(file_name)
if mime_type_guess is not None:
response = HttpResponse(fsock, mimetype=mime_type_guess[0])
response['Content-Disposition'] = 'attachment; filename=' + file_name
#response.write(file)
except IOError:
response = HttpResponseNotFound()
return response
EDIT: Updated the source according Ignacio Vazquez-Abrams comments:
import os.path
import mimetypes
#login_required
def serve_upload_files(request, file_url):
mimetypes.init()
try:
file_path = os.path.join(settings.UPLOAD_LOCATION, file_url)
#collapse possibly available up-level references
file_path = os.path.normpath(file_path)
#check if file path still begins with settings.UPLOAD_LOCATION, otherwise the user tampered around with up-level references in the url
#for example this input: http://127.0.0.1:8000/upload/..\test_upload.txt results having the user access to a folder one-level higher than the upload folder
#AND check if the common_prefix ends with a dir separator, Because although '/foo/barbaz' starts with '/foo/bar'
common_prefix = os.path.commonprefix([settings.UPLOAD_LOCATION, file_path])
if common_prefix == settings.UPLOAD_LOCATION and common_prefix.endswith(os.sep):
fsock = open(file_path,"r")
file_name = os.path.basename(file_path)
mime_type_guess = mimetypes.guess_type(file_name)
if mime_type_guess is not None:
response = HttpResponse(fsock, mimetype=mime_type_guess[0])
response['Content-Disposition'] = 'attachment; filename=' + file_name
else:
response = HttpResponseNotFound()
else:
print "wrong directory"
response = HttpResponseNotFound()
except IOError:
response = HttpResponseNotFound()
return response

Answer 1

A few tips:
Use os.path.join() to join the path together.
Use os.path.normpath() to get the actual path with no ".." references.
Use os.path.commonprefix() against UPLOAD_LOCATION and the generated path, and verify that the result starts with UPLOAD_LOCATION.
Make sure that UPLOAD_LOCATION ends with a dir separator.
TL;DR: Use os.path.