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.
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()
you can start the web.py by using this command
/usr/bin/python index.py > log.txt 2>&1 &
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