SOLVED:
Not a Web2py problem,but when i kill it, the port status remain in
TIME_WAIT,so another process cannot open the port again
If i wait about 30 seconds after kill the process,it restart without
problems
Il giorno lunedì 27 aprile 2015 23:44:29 UTC+2, tommasot ha scritto:
>
> I want run a simple script on web2py startup,that listen on a tcp port
>
> the script is the following "example.py". The file in the
> "applications/myapp/cron" folder
>
>
>
> *import SocketServer*
> *import logging*
>
> *class MyTCPHandler(SocketServer.BaseRequestHandler):*
>
>
> * # handle syslog event message*
> * def handle(self):*
> * self.logger = logging.getLogger("web2py.app.easylog")*
> * self.logger.setLevel(logging.DEBUG)*
> * messagelog = self.request.recv(1024).strip()*
> * self.logger.info <http://self.logger.info>(messagelog)*
>
> *if __name__ == "__main__":*
> * HOST, PORT = "0.0.0.0", 514*
> * servertcp = SocketServer.TCPServer((HOST, PORT), MyTCPHandler)*
>
>
> * servertcp.serve_forever()*
>
> i have created into the "cron" folder the crontab file with the following
> content:
> *@reboot root *applications/easylog/cron/example.py*
>
> *i run web2py with the following shell command "sudo python web2py.py -Y"*
>
> *The first time web2py is running everything works fine and i am able to
> connect with telnet on port 514*
>
> *If i kill web2py and restart it.... no more open port on 514*
>
> *The question is.... Why? *
>
>
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
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