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如何使用python SocketServer创建连接超时

[英]How to create connection timeout with python SocketServer

Good day! 美好的一天! I was writen simple server: 我写的是简单的服务器:

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. 如何设置连接超时。 I want when the client not send me data and is not active in 30 seconds, server will close connection. 我想当客户端不向我发送数据并且在30秒内未激活时,服务器将关闭连接。

PS sorry for my english. 对不起我的英语。

UPDATE 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()

If you use StreamRequestHandler instead of BaseRequestHandler, you just need to override the timeout variable there, and it will be set. 如果你使用StreamRequestHandler而不是BaseRequestHandler,你只需要覆盖那里的超时变量,它就会被设置。 If you want to learn how to do it yourself, just look at the SocketServer.py 如果您想自己学习如何操作,只需查看SocketServer.py即可

Here's an example, this will kill any connections that aren't done in 5 seconds: 这是一个例子,这将杀死5秒内未完成的任何连接:

#!/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. 这使用了python的socket settimeout()调用。

I don't think your alarm() solution will work with threading or forking. 我不认为你的alarm()解决方案可以用于线程或分叉。

Please look at it: 请看一下:

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. 如果要处理多个请求,则需要再次执行server.handle_request()。

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