1. You could read (and thus write) in smaller chunks, by giving read() an
argument indicating the number of bytes to read:
f = open(file) # The mode 'r' is default
for buf in read(1024*32):
response.write(buf)
f.close()
2. You could, instead, let Apache serve the files in encrypted form,
and sell the
user a decryption key.
On Mon, Nov 23, 2009 at 12:18 PM, 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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