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Home/ Questions/Q 8960063
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T15:35:20+00:00 2026-06-15T15:35:20+00:00

I want to start a web2py server so that it can be accessed externally

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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.

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  1. Editorial Team
    Editorial Team
    2026-06-15T15:35:21+00:00Added an answer on June 15, 2026 at 3:35 pm
    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

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