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Home/ Questions/Q 788663
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T21:24:10+00:00 2026-05-14T21:24:10+00:00

I have a Pylons controller action that needs to return a file to the

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I have a Pylons controller action that needs to return a file to the client. (The file is outside the web root, so I can’t just link directly to it.) The simplest way is, of course, this:

    with open(filepath, 'rb') as f:
        response.write(f.read())

That works, but it’s obviously inefficient for large files. What’s the best way to do this? I haven’t been able to find any convenient methods in Pylons to stream the contents of the file. Do I really have to write the code to read a chunk at a time myself from scratch?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T21:24:11+00:00Added an answer on May 14, 2026 at 9:24 pm

    I finally got it to work using the FileApp class, thanks to Chris AtLee and THC4k (from this answer). This method also allowed me to set the Content-Length header, something Pylons has a lot of trouble with, which enables the browser to show an estimate of the time remaining.

    Here’s the complete code:

    def _send_file_response(self, filepath):
        user_filename = '_'.join(filepath.split('/')[-2:])
        file_size = os.path.getsize(filepath)
    
        headers = [('Content-Disposition', 'attachment; filename=\"' + user_filename + '\"'),
                   ('Content-Type', 'text/plain'),
                   ('Content-Length', str(file_size))]
    
        from paste.fileapp import FileApp
        fapp = FileApp(filepath, headers=headers)
    
        return fapp(request.environ, self.start_response)
    
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