Run web.py as daemon

backgroundworker, python

Solution

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()

Problem

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.

Original source