Don't do it in the handle() method, pass the port number in (from this https://docs.python.org/2/library/socketserver.html#socketserver-tcpserver-example):
#!/usr/bin/env python
import SocketServer, argparse
class MyTCPHandler(SocketServer.BaseRequestHandler):
"""
The RequestHandler class for our server.
It is instantiated once per connection to the server, and must
override the handle() method to implement communication to the
client.
"""
def handle(self):
# self.server.server_address is a tuple (IP, port) the server is listening on
(host, port) = self.server.server_address
print 'port # is: {}'.format(port)
# self.request is the TCP socket connected to the client
self.data = self.request.recv(1024).strip()
print "{} wrote:".format(self.client_address[0])
print self.data
# just send back the same data, but upper-cased
self.request.sendall(self.data.upper())
if __name__ == "__main__":
parser = argparse.ArgumentParser()
parser.add_argument('-p', '--port', required=True, help='the TCP port to listen on')
args = parser.parse_args()
HOST, PORT = "localhost", int(args.port)
# Create the server, binding to localhost on port 9999
server = SocketServer.TCPServer((HOST, PORT), MyTCPHandler)
# Activate the server; this will keep running until you
# interrupt the program with Ctrl-C
server.serve_forever()
In this example, you must provide the port number as an argument when you start the program using the -p command line switch.