How to serve any static directory?

[Yann Salaün]

Hi all,

I'm working on an application where we want the following to happen:

1. The administrators upload zip archives containing web content (whole static websites) through admin interface

2. The archive is unzipped in the background to a random directory

3. The static content is served at a url that looks like http://<host>/site/<pk>/, with the index.html at the root, all the paths to css and js files, etc.

I know how to configure the route, but I was wondering what to put in the view. I have tried with STATIC_{ROOT,URL} and MEDIA_{ROOT,URL} but I feel it's not the right way.

In summary : is there a way to put something like `return serve_this_static_directory()` in the view? or is there any workaround there?

Thanks for your answers

Yann

[Florian Schweikert]

On 01/09/15 16:34, Yann Salaün wrote:
> In summary : is there a way to put something like `return
> serve_this_static_directory()` in the view? or is there any workaround
> there?

Not sure if I get you right, you want to serve static files with django?
Is there any reason why it's not possible to serve them with nginx/apache?

regards,
Florian

[Yann Salaün]

Hi,

The main reason I try to serve those static files with django is because I would like to use django to configure the route. 

Trying to clarify my question:

1. The url (http://<host>/site/<user-friendly-url>/) doesn't match the directory name (which could be anything, like `XYZ_html/`).

2. Administrators must be able to upload new archives, which should be extracted and served automatically. 

For those two reasons, I don't have any idea how to serve this content using the webserver, because how is it possible to configure the routes then?

Thank you for your answer

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[Remco Gerlich]

Do you use a webserver "in front" of Django, like Nginx or Apache?

What we do:

- In *development* (when DEBUG=True), use django.views.static.serve to serve a file: https://docs.djangoproject.com/en/1.8/ref/views/#django.views.static.serve

- In *production*, let Django do what it needs to do (like authorisation, or in your case the dynamic translation of the URL to a file path) and then send headers to the webserver to let it return the file, freeing up the Django worker.

Under Nginx, we have a bit of nginx.conf in the server {} part like:

    location /protected/ {

          internal;

          alias /the/directory/root/of/the/files/
    }

And we create a path that starts with /protected/. For Apache, you simply need the full absolute path.

Then do

    response = http.HttpResponse()

    response['X-Sendfile'] = full_absolute_path  # Apache

    response['X-Accel-Redirect'] = path_starting_with_protected  # Nginx

    response['Content-Type'] = ''  # So the webserver will decide it

    return respone

Hope this helps.

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