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

user_78374 0
Viewed 26434

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)

django
9 Answers

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

user_2214960

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

user_1043490

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

Ignacio_Vazquez-Abrams_208620

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
user_78489240

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)
user_139310610

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.

user_555620
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