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(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 the 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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