On Tuesday, November 1, 2011 5:46:57 PM UTC-4, John Duddy wrote: > > That lets me read the URL from my web2py code, right? Instead, I want > to serve the file via web2py to the browser. > > I suppose I could use this with response.stream, if I knew how to > parse the rest of the URL component below my controller....
response.stream() takes an open file object, so I suppose you could do
open('path/to/file', 'rb') to open the file for streaming.
Anthony
