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如何制作一个简单的 Python 回显服务器来记住客户端并且不会为每个请求创建一个新的套接字？必须能够支持并发访问。我希望能够使用此客户端或类似客户端连接一次并不断发送和接收数据：

import socket

sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
host = raw_input("Server hostname or ip? ")
port = input("Server port? ")
sock.connect((host,port))
while True:
    data = raw_input("message: ")
    sock.send(data)
    print "response: ", sock.recv(1024)

即服务器在端口 50000 上运行，使用上面的客户端我希望能够做到这一点：

me@mine:~$ client.py
Server hostname or ip? localhost
Server Port? 50000
message: testa
response: testa
message: testb
response: testb
message: testc
response: testc
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1 回答 1

94

您可以为每个客户端使用一个线程来避免阻塞，client.recv()然后使用主线程来监听新客户端。当一个连接时，主线程创建一个新线程，该线程只监听新客户端，并在它不说话 60 秒时结束。

import socket
import threading

class ThreadedServer(object):
    def __init__(self, host, port):
        self.host = host
        self.port = port
        self.sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
        self.sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
        self.sock.bind((self.host, self.port))

    def listen(self):
        self.sock.listen(5)
        while True:
            client, address = self.sock.accept()
            client.settimeout(60)
            threading.Thread(target = self.listenToClient,args = (client,address)).start()

    def listenToClient(self, client, address):
        size = 1024
        while True:
            try:
                data = client.recv(size)
                if data:
                    # Set the response to echo back the recieved data 
                    response = data
                    client.send(response)
                else:
                    raise error('Client disconnected')
            except:
                client.close()
                return False

if __name__ == "__main__":
    while True:
        port_num = input("Port? ")
        try:
            port_num = int(port_num)
            break
        except ValueError:
            pass

    ThreadedServer('',port_num).listen()

客户端在 60 秒不活动后超时，必须重新连接。查看client.settimeout(60)函数中的行ThreadedServer.listen()

于 2014-05-23T11:39:56.850 回答