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.
just works fine but the log message will point you to an invalid address:
alternatively you can use ethernet interface ip but it will not listen also on localhost