如何保持套接字打开，直到客户端关闭它？

Question

我有简单的python服务器和客户端。

服务器：

import SocketServer
import threading


class MyTCPHandler(SocketServer.BaseRequestHandler):
    def handle(self):
        self.data = self.request.recv(1024).strip()
        print str(self.client_address[0]) + " wrote: "
        print self.data
        self.request.send(self.data.upper())


if __name__ == "__main__":
    HOST, PORT = "localhost", 3288
    server = SocketServer.TCPServer((HOST, PORT), MyTCPHandler)
    server.serve_forever()

客户：

import socket
import sys
from time import sleep

HOST, PORT = "localhost", 3288
data = "hello"

sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

try:
    sock.connect((HOST, PORT))
    sock.send(data + "\n")
    received = sock.recv(1024)

    sleep(10)

    sock.send(data + "\n")
    received = sock.recv(1024)

    sleep(10)

    sock.send(data + "\n")
    received = sock.recv(1024)

finally:
    sock.close()

这是我得到的输出：

服务器：

>python server.py
127.0.0.1 wrote:
hello

客户：

>python client.py
Traceback (most recent call last):
  File "client.py", line 18, in <module>
    received = sock.recv(1024)
socket.error: [Errno 10053] An established connection was aborted by the software in your host machine

我也在Linux机器上尝试过。服务器仅收到一条消息，然后第二条消息的recv语句出现错误。我刚刚开始在python上学习网络，但是我认为服务器由于某种原因正在关闭套接字。我该如何纠正？

Answer 1

MyTcpHandler为每个连接创建一个对象，并handle调用该对象以处理客户端。handle返回时连接将关闭，因此您必须在方法中处理来自客户端的完整通信handle：

class MyTCPHandler(SocketServer.BaseRequestHandler):
    def handle(self):
        while 1:
            self.data = self.request.recv(1024)
            if not self.data:
                break
            self.data = self.data.strip()
            print str(self.client_address[0]) + " wrote: "
            print self.data
            self.request.send(self.data.upper())

注意：当客户端关闭连接时recv返回''，因此我移动了.strip()之后，recv因此没有错误警报，因为客户端仅发送空白。