我遵循了一位名为 DrapsTV 的 YouTube 用户的教程。本教程是用 Python 2.7 编写的,并使用 UDP 进行网络聊天。我将其转换为 Python 3 并让一切正常工作。然而,线程的设置方式是我必须发送一条消息(或按 Enter 键,这是一条空白消息)来刷新和接收来自其他客户端的任何消息。这是视频,以防您可能需要:https://www.youtube.com/watch?v=PkfwX6RjRaI
这是我的服务器代码:
from socket import *
import time
address = input("IP Address: ")
port = input("Port: ")
clients = []
serversock = socket(AF_INET, SOCK_DGRAM)
serversock.bind((address, int(port)))
serversock.setblocking(0)
quitting = False
print("Server is up and running so far.")
while not quitting:
try:
data, addr = serversock.recvfrom(1024)
if "Quit" in str(data):
quitting = True
if addr not in clients:
clients.append(addr)
print(time.ctime(time.time()) + str(addr) + ": :" + str(data.decode()))
for client in clients:
serversock.sendto(data, client)
except:
pass
serversock.close()
这是我的客户端代码:
from socket import *
import threading
import time
tLock = threading.Lock()
shutdown = False
def receiving(name, sock):
while not shutdown:
try:
tLock.acquire()
while True:
data, addr = sock.recvfrom(1024)
print(str(data.decode()))
except:
pass
finally:
tLock.release()
address = input("IP Address: ")
port = 0
server = address, 6090
clientsock = socket(AF_INET, SOCK_DGRAM)
clientsock.setsockopt(SOL_SOCKET, SO_REUSEADDR, 1)
clientsock.bind((address, int(port)))
clientsock.setblocking(0)
rT = threading.Thread(target=receiving, args=("RecvThread", clientsock))
rT.start()
nick = input("How about we get you a nickname: ")
message = input(nick + "> ").encode()
while message != "q":
if message != "":
clientsock.sendto(nick.encode() + "> ".encode() + message, server)
tLock.acquire()
message = input(nick + "> ").encode()
tLock.release()
time.sleep(0.2)
shutdown = True
rT.join()
clientsock.close()
最佳答案
@furas 善意地为我解释了我的问题:这不是我的接收方法有缺陷(例如我的线程或函数),而是输入调用阻止了客户端接收任何内容。因此,为了解决这个问题,我或其他遇到此问题的人需要找到一种方法来在按下某个按钮时调用输入,以便除非您打字,否则您可以接收消息或数据。
关于python - (Python) 如何在不刷新的情况下实时接收消息?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40569643/