web2py - allow external access - how?

Question

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

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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.

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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.

Answer 1

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

Answer 2

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.