How to create connection timeout with python SocketServer

Question

Good day! I was writen simple server:

class SingleTCPHandler(SocketServer.BaseRequestHandler):

    def handle(self):
        data = self.request.recv(1024) 
        self.request.close()

class SimpleServer(SocketServer.ThreadingMixIn, SocketServer.TCPServer):

    daemon_threads = True

    allow_reuse_address = True

    def __init__(self, server_address, RequestHandlerClass):
        SocketServer.TCPServer.__init__(self, server_address, RequestHandlerClass)


def running():
    server = SimpleServer((settings.host, settings.port), SingleTCPHandler)
    try:
        server.serve_forever()
    except KeyboardInterrupt:
        sys.exit(0)

How to set connection timeout. I want when the client not send me data and is not active in 30 seconds, server will close connection.

P.S. sorry for my english.

UPDATE

#!/usr/bin/env python
# -*- coding: utf8 -*-

import sys
import time

import SocketServer
import datetime
import settings
import os
from signal import SIGTERM, SIGCHLD, signal, alarm
import socket
import subprocess
from threading import Thread
import MySQLdb
import re

class SingleTCPHandler(SocketServer.BaseRequestHandler):
    "One instance per connection.  Override handle(self) to customize action."
    def handle(self):
        alarm(30)
        data = self.request.recv(1024) 
        # Some code
        self.request.close()


class SimpleServer(SocketServer.ForkingMixIn, SocketServer.TCPServer):

    daemon_threads = True
    allow_reuse_address = True


    def __init__(self, server_address, RequestHandlerClass):
        SocketServer.TCPServer.__init__(self, server_address, RequestHandlerClass)




def running():
    server = SimpleServer((settings.host, settings.port), SingleTCPHandler)
    try:
        server.serve_forever()
    except KeyboardInterrupt:
        sys.exit(0)


def deamonize(stdout='/dev/null', stderr=None, stdin='/dev/null', pidfile=None, startmsg='started with pid %s'):
    try:
        pid = os.fork()
        if (pid > 0):
            sys.exit(0)
    except OSError, e:
        sys.stderr.write("fork #1 failed: (%d) %s\n" % (e.errno, e.strerror))
        sys.exit(1)

    os.chdir(settings.place)
    os.umask(0)
    os.setsid()

    try:
        pid = os.fork()
        if (pid > 0):
            sys.exit(0) 
    except OSError, e:
        sys.stderr.write("fork #2 failed: (%d) %s\n" % (e.errno, e.strerror))
        sys.exit(1)

    if (not stderr):
        stderr = stdout

        print stdin, stdout, stderr
        si = file(stdin, 'r')
        so = file(stdout, 'a+')
        se = file(stderr, 'a+', 0)
        pid = str(os.getpid())
        sys.stderr.write("\n%s\n" % startmsg % pid)
        sys.stderr.flush()
    if pidfile: file(pidfile, 'w+').write("%s\n" % pid)

    os.dup2(si.fileno(), sys.stdin.fileno())
    os.dup2(so.fileno(), sys.stdout.fileno())
    os.dup2(se.fileno(), sys.stderr.fileno())

def startstop(stdout='/dev/null', stderr=None, stdin='/dev/null', pidfile='pid.txt', startmsg='started with pid %s'):
    if len(sys.argv) > 1:
        action = sys.argv[1]
        try:
            pf = open(pidfile)
            pid = int(pf.read().strip())
            pf.close()
        except IOError:
            pid = None
        if ((action == 'stop') or (action == 'restart')):
            if (not pid):
                mess = "Не могу остановить, pid файл '%s' отсутствует.\n"
                sys.stderr.write(mess % pidfile)
                sys.exit(1)
            try:
                while 1:
                    os.kill(pid, SIGTERM)
                    time.sleep(1)
            except OSError, err:
                err = str(err)
                if err.find("No such process") > 0:
                    os.remove(pidfile)
                    if 'stop' == action:
                        sys.exit(0)
                    action = 'start'
                    pid = None
                else:
                    print str(err)
                    sys.exit(1)
        if ('start' == action):
            if (pid):
                mess = "Старт отменен — pid файл '%s' существует.\n"
                sys.stderr.write(mess % pidfile)
                sys.exit(1)
            deamonize(stdout, stderr, stdin, pidfile, startmsg)
            return
    print "Синтакс запуска: %s start|stop|restart" % sys.argv[0]
    sys.exit(2)

if (__name__ == "__main__"):
    startstop(stdout=settings.log, pidfile=settings.pid)
    running()

Answer 1

If you use StreamRequestHandler instead of BaseRequestHandler, you just need to override the timeout variable there, and it will be set. If you want to learn how to do it yourself, just look at the SocketServer.py

Here's an example, this will kill any connections that aren't done in 5 seconds:

#!/usr/bin/env python
import SocketServer

class myHandler(SocketServer.StreamRequestHandler):
    timeout = 5
    def handle(self):
        recvdata = ""
        while True:
            tmp = self.request.recv(16384)
            recvdata = recvdata + tmp.strip()
            if (len(tmp) < 16384):
                break;
        self.request.send("Received: {0}".format(recvdata))

class myApp(SocketServer.TCPServer):

    def __init__(self):
        SocketServer.TCPServer.__init__(self, ("localhost", 5555), myHandler)
        print self.server_address
        try:
            self.serve_forever()
        except KeyboardInterrupt:
            print "Got keyboard interrupt, shutting down"
            self.shutdown()

if __name__ == "__main__":
    app = myApp()

This uses the python's socket settimeout() call.

I don't think your alarm() solution will work with threading or forking.

Answer 2

Please look at it:

import sys
import SocketServer

class SingleTCPHandler(SocketServer.BaseRequestHandler):
    def handle(self):
        data = self.request.recv(1024)
        self.request.close()

class SimpleServer(SocketServer.ThreadingMixIn, SocketServer.TCPServer):

    timeout = 30

    daemon_threads = True
    allow_reuse_address = True

    def __init__(self, server_address, RequestHandlerClass):
        SocketServer.TCPServer.__init__(self, server_address, RequestHandlerClass)

    def handle_timeout(self):
        print 'Timeout!'


def running():
    server = SimpleServer(('localhost', 6666), SingleTCPHandler)
    try:
        #server.serve_forever()
        server.handle_request()
    except KeyboardInterrupt:
        sys.exit(0)

if __name__ == '__main__':
    running()

# vim: filetype=python syntax=python expandtab shiftwidth=4 softtabstop=4 encoding=utf8

If you want to handle more than one request you need to execute server.handle_request() again.