用Python创建一个简单的聊天应用程序（套接字）

我正在尝试使用套接字（python）创建一个简单的聊天应用程序。其中，客户端可以向服务器发送消息，服务器只需将消息广播给所有其他客户端（发送消息的客户端除外）

客户端有两个线程，它们永远运行

send

：send只是将cleints消息发送到服务器

receive

：从服务器接收消息

服务器还有两个线程，它们将永远运行

accept\u cleint

：接受来自客户端的传入连接

broadcast\u usr

：接受来自客户端的消息并将其广播到所有其他客户端

但是我得到了错误的输出（请参考下图）。所有线程都假定一直处于活动状态，但有时客户端可以发送消息，有时则不能。比如说Tracey发了4次“hi”，但没有广播，当John说了2次“bye”，然后1次信息被广播。服务器上似乎有一些线程同步问题，我不确定。请告诉我怎么了

下面是代码

chat_client.py

import socket, threading

def send():
    while True:
        msg = raw_input('\nMe > ')
        cli_sock.send(msg)

def receive():
    while True:
        sen_name = cli_sock.recv(1024)
        data = cli_sock.recv(1024)

        print('\n' + str(sen_name) + ' > ' + str(data))

if __name__ == "__main__":   
    # socket
    cli_sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

    # connect
    HOST = 'localhost'
    PORT = 5023
    cli_sock.connect((HOST, PORT))     
    print('Connected to remote host...')
    uname = raw_input('Enter your name to enter the chat > ')
    cli_sock.send(uname)

    thread_send = threading.Thread(target = send)
    thread_send.start()

    thread_receive = threading.Thread(target = receive)
    thread_receive.start()

chat_server.py

import socket, threading

def accept_client():
    while True:
        #accept    
        cli_sock, cli_add = ser_sock.accept()
        uname = cli_sock.recv(1024)
        CONNECTION_LIST.append((uname, cli_sock))
        print('%s is now connected' %uname)

def broadcast_usr():
    while True:
        for i in range(len(CONNECTION_LIST)):
            try:
                data = CONNECTION_LIST[i][1].recv(1024)
                if data:
                    b_usr(CONNECTION_LIST[i][1], CONNECTION_LIST[i][0], data)
            except Exception as x:
                print(x.message)
                break

def b_usr(cs_sock, sen_name, msg):
    for i in range(len(CONNECTION_LIST)):
        if (CONNECTION_LIST[i][1] != cs_sock):
            CONNECTION_LIST[i][1].send(sen_name)
            CONNECTION_LIST[i][1].send(msg)

if __name__ == "__main__":    
    CONNECTION_LIST = []

    # socket
    ser_sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

    # bind
    HOST = 'localhost'
    PORT = 5023
    ser_sock.bind((HOST, PORT))

    # listen    
    ser_sock.listen(1)
    print('Chat server started on port : ' + str(PORT))

    thread_ac = threading.Thread(target = accept_client)
    thread_ac.start()

    thread_bs = threading.Thread(target = broadcast_usr)
    thread_bs.start()

---

好吧，我之前在评论中撒了谎，对不起。问题实际上是服务器上的

broadcast\u usr（）

函数。它在

recv（）

方法中阻塞，并阻止除当前所选用户之外的所有用户在通过

for

循环时一次讲话。为了解决这个问题，我修改了server.py程序，为它接受的每个客户端连接生成一个新的broadcast_usr线程。我希望这有帮助

import socket, threading

def accept_client():
    while True:
        #accept    
        cli_sock, cli_add = ser_sock.accept()
        uname = cli_sock.recv(1024)
        CONNECTION_LIST.append((uname, cli_sock))
        print('%s is now connected' %uname)
        thread_client = threading.Thread(target = broadcast_usr, args=[uname, cli_sock])
        thread_client.start()

def broadcast_usr(uname, cli_sock):
    while True:
        try:
            data = cli_sock.recv(1024)
            if data:
                print "{0} spoke".format(uname)
                b_usr(cli_sock, uname, data)
        except Exception as x:
            print(x.message)
            break

def b_usr(cs_sock, sen_name, msg):
    for client in CONNECTION_LIST:
        if client[1] != cs_sock:
            client[1].send(sen_name)
            client[1].send(msg)

if __name__ == "__main__":    
    CONNECTION_LIST = []

    # socket
    ser_sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

    # bind
    HOST = 'localhost'
    PORT = 5023
    ser_sock.bind((HOST, PORT))

    # listen    
    ser_sock.listen(1)
    print('Chat server started on port : ' + str(PORT))

    thread_ac = threading.Thread(target = accept_client)
    thread_ac.start()

    #thread_bs = threading.Thread(target = broadcast_usr)
    #thread_bs.start()

---

我试着避开你说的“阿提尼什”这只虫子。客户端将被询问一次用户名，此“uname”将包含在要发送的数据中。请参阅我对“发送”函数所做的操作

为了便于可视化，我在所有收到的消息中添加了一个“\t”

import socket, threading

def send(uname):
    while True:
        msg = raw_input('\nMe > ')
        data = uname + '>' + msg
        cli_sock.send(data)

def receive():
    while True:
        data = cli_sock.recv(1024)
        print('\t'+ str(data))

if __name__ == "__main__":   
    # socket
    cli_sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

    # connect
    HOST = 'localhost'
    PORT = 5023

    uname = raw_input('Enter your name to enter the chat > ')

    cli_sock.connect((HOST, PORT))     
    print('Connected to remote host...')


    thread_send = threading.Thread(target = send,args=[uname])
    thread_send.start()

    thread_receive = threading.Thread(target = receive)
    thread_receive.start()

您还必须相应地修改服务器代码

server.py

import socket, threading

def accept_client():
    while True:
        #accept    
        cli_sock, cli_add = ser_sock.accept()
        CONNECTION_LIST.append(cli_sock)
        thread_client = threading.Thread(target = broadcast_usr, args=[cli_sock])
        thread_client.start()

def broadcast_usr(cli_sock):
    while True:
        try:
            data = cli_sock.recv(1024)
            if data:
               b_usr(cli_sock, data)
         except Exception as x:
            print(x.message)
            break

def b_usr(cs_sock, msg):
    for client in CONNECTION_LIST:
        if client != cs_sock:
            client.send(msg)

if __name__ == "__main__":    
    CONNECTION_LIST = []

    # socket
    ser_sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)

    # bind
    HOST = 'localhost'
    PORT = 5023
    ser_sock.bind((HOST, PORT))

    # listen    
    ser_sock.listen(1)
    print('Chat server started on port : ' + str(PORT))

    thread_ac = threading.Thread(target = accept_client)
    thread_ac.start()

---

服务器端发生的变化是：连接的用户和发言的用户不再被看到。如果你的目的是联系客户，我不知道这是否意味着什么。如果您想通过服务器严格监控客户端，可能还有另一种方法。

我认为问题在于您的客户端线程循环，尽管您的服务器也需要能够处理客户端断开连接。仅供参考，Twisted是一个用于制作线程服务器的Python库。您的代码运行良好，但存在一个小错误。发送者必须发送2条消息才能广播。客户端的

sen\u name=cli\u sock.recv（1024）

和

data=cli\u sock.recv（1024）

都在接收发送方名称和消息。请参考下面的屏幕截图，服务器代码第17行有缩进错误。它只需要向左移动1个字符。