I want to start a web2py server so that it can be accessed externally to the hosting server.
I've read this http://web2py.com/books/default/chapter/29/03
By default, web2py runs its web server on 127.0.0.1:8000 (port 8000 on localhost), but you can run it on any available IP address and port. You can query the IP address of your network interface by opening a command line and typing ipconfig on Windows or ifconfig on OS X and Linux. From now on we assume web2py is running on localhost (127.0.0.1:8000). Use 0.0.0.0:80 to run web2py publicly on any of your network interfaces.
but I can't find how to "Use 0.0.0.0:80" ? There doesn't seem to be a command line argument which would do that.
Thanks
EDIT: I should say the server in question does not have a GUI - I'm aware there's some sort GUI based admin facilties for web2py but that's out of the question here.
EDIT2: Just in case this is not clear (and on the offchance it makes any difference - which I doubt) I'm running the server like this :
sudo python web2py.py
not via wsgi/apache or the like.
python web2py.py --ip 0.0.0.0
just works fine but the log message will point you to an invalid address:
please visit:
http://0.0.0.0:8000
alternatively you can use ethernet interface ip but it will not listen also on localhost
What may help you is the fact that you can select the public ip when the server gui pops up asking for the admin password.
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