Re: Allowing external access to web2py server ?

[Massimo Di Pierro]

python web2py.py -a passwd -i 0.0.0.0 -p 80

On Friday, 7 December 2012 03:21:55 UTC-6, Richard Shea wrote:

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 does that ?

I'm aware of the command line arg to change the port but how do I allow access from all addresses.

thanks

Richard.

[Richard Shea]

Thank you Massimo.

[sebas mora]

Help please! Im trying to access my web2py server on a Rpi from another device to run an application that is on the server. From local host it runs and executes fine, no problem, but i just cant connect to it from other devices. 

I do what is said in this message and I get the following error: 

ERROR:Rocker.Errors.Port80:Socket 192.168.1.172:80 in use by other process and it wont share.

What am I doing wrong? How can i get around this? 

 Please help!!

[Massimo Di Pierro]

You have to find out which other process is using port 80 and kill it.

You can do it with netstat:

netstat -a -b