当对等体持续监听传入连接(新对等体)并通过终端(用户输入)向其他对等体发送命令时,构建p2p系统。我总是在寻找新的同伴时,很难从键盘上寻找用户输入。
print 'Listening...'
while not shutdown:
while sys.stdin in select.select([sys.stdin], [], [], 0)[0]: #look for keyboard input... not working
line = sys.stdin.readline()
if line:
send_message(line)
else: # an empty line means stdin has been closed
print('eof')
exit(0)
try: # listen for other peers
clientsock,clientaddr = mySocket.accept()
print 'Incoming connection from', clientaddr
clientsock.settimeout(None)
t = threading.Thread( target = HandlePeer, args = [clientsock] )
t.start()
except KeyboardInterrupt:
print "shutting down"
shutdown = True
continue
except Exception,e:
print 'error in peer connection %s %s' % (Exception,e)
mySocket.close()
HandlePeer检查来自新连接对等方的传入消息。我只需要一种发送消息的方式。
答案 0 :(得分:1)
简短的回答是你需要使用curses。
比调用input()
并收到回复要困难得多,但这是你需要的。有一个名为Curses Programming with Python的好资源,这是最好的起点。