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Home/ Questions/Q 6995455
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T20:01:41+00:00 2026-05-27T20:01:41+00:00

I’ve been using a method for serving downloads but since it was not secure

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I’ve been using a method for serving downloads but since it was not secure i decided to change that . ( the method was a link to the original file in storage , but the risk was that everyone with the link could have downloaded the file ! ) so i now serve the file via my views , that way only users with permission can download the file , but i’m noticing a high load on server while there is many simultaneous download requests for the files. here’s part of my code that handles downloads for users ( Consider the file is an image )

    image = Image.open ("the path to file")
    response = HttpResponse(mimetype = 'image/png' )
    response['Content-Disposition'] = 'attachment: filename=%s.png' % filename
    image.save(response , "png")
    return response  

is there any better ways for serving files while keeping the security and lowering server side load ?
thanks in advance 🙂

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T20:01:42+00:00Added an answer on May 27, 2026 at 8:01 pm

    Your opening of the image loads it in memory and this is what causes the increase in load under heavy use. As posted by Martin the real solution is to serve the file directly.

    Here is another approach, which will stream your file in chunks without loading it in memory.

    import os
    import mimetypes
    
    from wsgiref.util import FileWrapper
    
    from django.http import StreamingHttpResponse
    
    
    def download_file(request):
        the_file = "/some/file/name.png"
        filename = os.path.basename(the_file)
        chunk_size = 8192
        response = StreamingHttpResponse(
            FileWrapper(
                open(the_file, "rb"),
                chunk_size,
            ),
            content_type=mimetypes.guess_type(the_file)[0],
        )
        response["Content-Length"] = os.path.getsize(the_file)
        response["Content-Disposition"] = f"attachment; filename={filename}"
        return response
    
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