Run web.py as daemon

Question

I have a simple web.py program to load data. In the server I don't want to install apache or any webserver.

I try to put it as a background service with http://www.jejik.com/articles/2007/02/a_simple_unix_linux_daemon_in_python/

And subclassing: (from http://www.jejik.com/files/examples/daemon.py)

class Daemon:
    def start(self):
        """
        Start the daemon
        """
        ... PID CHECKS....

        # Start the daemon
        self.daemonize()
        self.run()
#My code
class WebService(Daemon):
        def run(self):
            app.run()

if __name__ == "__main__":
    if DEBUG:
        app.run()
    else:
        service = WebService(os.path.join(DIR_ACTUAL,'ElAdministrador.pid'))
        if len(sys.argv) == 2:
            if 'start' == sys.argv[1]:
                service.start()
            elif 'stop' == sys.argv[1]:
                service.stop()
            elif 'restart' == sys.argv[1]:
                service.restart()
            else:
                print "Unknown command"
                sys.exit(2)
            sys.exit(0)
        else:
            print "usage: %s start|stop|restart" % sys.argv[0]
            sys.exit(2)

However, the web.py software not load (ie: The service no listen)

If I call it directly (ie: No using the daemon code) work fine.

Answer 1

I finally find the problem.

Web.py accept from command-line the optional port number:

python code.py 80

And the script also take input from the command-line:

python WebServer start

then web.py try to use "start" as port number and fail. I don't see the error because was in the bacground.

I fix this with a mini-hack:

if __name__ == "__main__":
    if DEBUG:
        app.run()
    else:
        service = WebService(os.path.join(DIR_ACTUAL,'ElAdministrador.pid'))
        if len(sys.argv) == 2:
            if 'start' == sys.argv[1]:
                sys.argv[1] = '8080'
                service.start()