파이썬 SocketServer와 연결 시간 제한을 만드는 방법

Question

좋은 하루! 내가 간단한 서버로 서면했다 :

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) 

연결 시간 제한을 설정하는 방법. 클라이언트가 데이터를 보내지 않고 30 초 안에 활성화되지 않으면 서버가 연결을 닫습니다.

P. 내 영어로 미안해.

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

그것을보고하십시오 :

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 

을 당신이()를 다시 server.handle_request을 실행할 필요 이상의 요청을 처리합니다.

Answer 2

BaseRequestHandler 대신 StreamRequestHandler를 사용하는 경우 해당 시간 초과 변수를 재정 의하여해야 설정됩니다.

#!/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() 

이가 사용 : 당신이 직접 작업을 수행하는 방법을 배우고 싶다면, 바로 여기에 예입니다,이 5 초에서 수행되지 않은 연결을 죽일 것 SocketServer.py

보고 파이썬의 socket settimeout() 호출.

귀하의 alarm() 솔루션이 스레딩 또는 분기와 함께 작동하지 않을 것이라고 생각합니다.