Hi Adrian,
if you use Apache, you can use mod_xsendfile[1] (other servers have
similar functionalities). Then you write somethink like this in your
view:
import mimetypes
[...]
mimetype, encoding = mimetypes.guess_type(path)
response = HttpResponse("", content_type=mimetype)
response['X-Sendfile'] = absolute_path_to_file
response['Content-Disposition'] = 'attachment; file=%s' %
os.path.split(path)[-1]
return response
Apache will see the X-Sendfile header and handle the rest for you.
If this is a high volume site, you'd probably better use a lightweight
reverse proxy like nginx[1], [2].
Regards,
Benjamin
[1] http://wiki.nginx.org/Main
[2] http://wiki.nginx.org/NginxXSendfile
On 23 Nov., 18:18, Adrián Ribao <ari...@gmail.com> wrote:
> Hello,
>
> I have created a website where you can buy and download some files.
> For security reasons all the files are served through a view, where
> comprobations are made in order to assure that the user bought the
> product.
>
> The problem is that the files are at least 400Mb and some of them are
> nearly 1Gb. Using this view is inefficient since all the content is
> loaded into memory and it kills the server:
>
> type, encoding = mimetypes.guess_type(file)
> name = os.path.basename(file)
> response = HttpResponse(mimetype=type)
> response['Content-Disposition'] = 'attachment; filename=%s' %
> (smart_str(name),)
>
> f = open(file, 'r')
> response.write(f.read())
> f.close()
> return response
>
> How could I solve this problem?
>
> Regards,
>
> Adrian.
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