Rich Lovely wrote:

I'd like to draw your attention to my original message: "a google
search is similarly lacking in relevant results."  Shouldn't it be
clear from how easy it is to find SimpleHttp that it clearly /isn't/
what I'm looking for? Perhaps I should have mentioned that, but I
thought I'd made it clear I'd tried a websearch.

Yes, you should have mentioned that SimpleHttp was not the one-liner you were looking for. That would have been useful to know :)

Your request was for a one-liner web server, and you claimed to have done a search which came up with nothing relevant. Since a simple search did in fact come up with an apparently relevant one-liner, the most obvious conclusions were that your search skills are very lousy, or that you were mistaken (i.e. lying) about having attempted to search first. Coming from an apparent first-time poster with no reputation I was aware of, neither would have surprised me. I'm glad that neither is actually the case.


The code I remember was a Socket-based webserver, written in one -
albeit rather long - line.

That would have been useful information to have mentioned at the time.

For what little it's worth, here's a 15 line socket-based web server:


import socket
template = ('HTTP/1.0 200 OK\n\n<html><head><title>Welcome %s!</title>'
  '</head><body><h1>Header...</h1>...and body.<br>Your request was: '
  '"%s"</body></html>')
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
sock.bind(('', 8080))
sock.listen(1)
print "Listening on port 8080"
while True:
    client, address = sock.accept()
    cfile = client.makefile('rw', 0)
    cfile.write(template % (str(address), cfile.readline().strip()))
    cfile.close()
    client.close()


Save this as "web.py", run "python web.py", and then in your browser go to http://localhost:8080

Turning this into a one-liner is left as an exercise. This may be useful:

http://pauldotcom.com/2011/10/python-one-line-shell-code.html



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--
Steven
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