How to return static files passing through a view in django?

Question

I need to return css files and js files according to specific logic. Clearly, static serve does not perform what I need. I have a view, whose render method uses logic to find the proper file, but then I have to return it. Technically, I can just read the file and stuff it into a HttpResponse object with the proper mime type, but I was wondering if there was a better strategy. (like fpassthru() in php)

Answer 1

This is what I used:

test_file = open('/home/poop/serve/test.pdf', 'rb')
response = HttpResponse(content=test_file)
response['Content-Type'] = 'application/pdf'
response['Content-Disposition'] = 'attachment; filename="%s.pdf"' \
                                  % 'whatever'
return response

Answer 2

What webserver software are you using?

At least for Apache and NginX, there is a module enabling you to use the X-SendFile HTTP header. The NginX website says Lighty can do this, too.

In your wrapper view:

...

abspath = '/most_secret_directory_on_the_whole_filesystem/protected_filename.css'

response = HttpResponse()
response['X-Sendfile'] = abspath

response['Content-Type'] = 'mimetype/submimetype'
# or let your webserver auto-inject such a header field
# after auto-recognition of mimetype based on filename extension

response['Content-Length'] = <filesize>
# can probably be left out if you don't want to hassle with getting it off disk.
# oh, and:
# if the file is stored via a models.FileField, you just need myfilefield.size

response['Content-Disposition'] = 'attachment; filename=%s.css' \
    % 'whatever_public_filename_you_need_it_to_be'

return response

Then you can connect the view via http://mysite.com/url_path/to/serve_hidden_css_file/.

You can use it anytime you need to do something upon a file being requested that should not be directly accessible to users, like limiting who can access it, or counting requests to it for stats, or whatever.

For Apache: http://tn123.ath.cx/mod_xsendfile/
For NginX: http://wiki.nginx.org/NginxXSendfile

Answer 3

Why don't you use Django staticfiles inside your view

from django.contrib.staticfiles.views import serve

...
def view_function(request):
   return serve(request, 'absolute_path_to_file_name')

Answer 4

Why not return an HttpResponseRedirect to the location of the correct static file?

Answer 5

Serving files directly from a view is very slow. If you are looking for normal file serving see this question: Having Django serve downloadable files

To very easily serve files through a view (for debug purposes, for example) keep reading.

# In your urls.py:
url(r'^test-files/(?P<name>.+)/$', views.test_files, name='test_files'),

# In your views.py:
from django.http.response import HttpResponse
from django.views.decorators.csrf import csrf_exempt

@csrf_exempt # (Allows file download with POST requests, can be omitted)
def test_files(request, name):
    if name == "myxml":
        fsock = open("/djangopath/data/static/files/my.xml", "rb")
        return HttpResponse(fsock)

This allows you to download the file from: http://127.0.0.1:8080/app/test-files/myxml/

Answer 6

Pass an iterator (such as the result of open()) to the HttpResponse constructor.

Answer 7

you can use below code in your view:
Note:in this function I return images but you can return every thing based your need and set your context_type

from django.http import HttpResponse,Http404
import os

def img_finder(request, img_name):
    try:
        with open(os.path.dirname(os.path.abspath(__file__)) + '/static/img/' + img_name, 'rb') as f:
            return HttpResponse(f.read(), content_type="image/jpeg")
    except IOError:
        raise Http404

Answer 8

Here the most simple and efficient way to do this.

app/urls.py

from django.urls import re_path
from app import views

urlpatterns = [
    re_path(r'^(?P<public_url>.*)$', views.public, name="public"),
]

Warning : put the URL pattern at the end

app/views.py

import os
from django.conf import settings
from django.views.static import serve

def public(request, public_url):
    public_folder = os.path.join(str(settings.BASE_DIR), 'folder_path')
    return serve(request, public_url, document_root=public_folder)

Answer 9

It should be wasteful to use django to serve static content (not to mention, several orders of magnitude slower).

I'd rather convert the view into a context processor and use the variables in templates to find what blocks to include.