線程和聯網

Question

我不知道如何讓有多人連接到我的Python服務器，以便能夠輸入消息，並立即看到它，而無需其他人按下輸入來查看它。我將如何在這裏使用線程？

服務器：

import socket, thread 

def reactor(sock,client,data): 
     try: 
      sock.sendto(data.encode('utf-8'),client) 
      sock.close() 
     except Exception,e : 
      print(e) 
      sock.close() 

def Main(): 
    host = '25.4.221.63' 
    port = 5000 

    clients = [] 


    s = socket.socket(socket.AF_INET,socket.SOCK_DGRAM) 
    s.bind((host,port)) 

    print("Server Started") 
    while True: 
     data,addr=s.recvfrom(1024) 
     if addr not in clients: 
      clients.append(addr) 
     data = data.decode('utf-8') 
     print("Message From: " + str(addr)) 
     print("From connected user: " + data) 
     print("Sending: " + data) 
     for client in clients: 
      try: 
       thread.start_new_thread(reactor,(s,client,data)) 
      except Exception,e : 
       print(e) 
       s.close() 



if __name__ == '__main__': 
    Main() 

客戶：

import socket 

def Main(): 
    alias = input("ScreenName:") 
    inp = input("IP:") 

    host =inp 
    port = 5001 

    server = ('25.4.221.63',5000) 

    s = socket.socket(socket.AF_INET,socket.SOCK_DGRAM) 
    s.bind((host,port)) 

    message = input("-> ") 

    while message != 'q': 
     s.sendto(message.encode('utf-8'), server) 
     data,addr = s.recvfrom(1024) 
     data = data.decode('utf-8') 
     print(data) 
     message = input(alias + ' -> ') 
    s.close() 


if __name__ == '__main__': 
    Main() 

Answer 1

避免線程;使用select。

您應該使用select或epoll來同時等待來自套接字或標準輸入的輸入。如果select表示套接字已準備好閱讀，請閱讀您的消息。如果stdin準備好讀取，請閱讀用戶輸入。這將需要您自己實施input。