python udp客户端超时机制

Question

如果Server socket中生成的随机数小于4，我的client socket会暂停接收数据。我需要设置超时机制让client socket检测到“超时”，然后它会继续发送消息。在我运行服务器套接字然后运行客户端套接字后，出现如下错误消息:

Traceback (most recent call last):
File "E:\Studying\Python workspace\Client\src\Client.py", line 34, in <module>
data , addr = client.recvfrom(1024)
socket.timeout: timed out

服务器套接字:

import random
from socket import *
serverSocket = socket(AF_INET , SOCK_DGRAM)
serverSocket.bind(('', 15000))

while True:

rand = random.randint(0, 10)
message , address = serverSocket.recvfrom (1024)
message = message.upper()
print("received message: ", message)
print("echo to address: ", address)
print(rand)

if rand < 4:
    continue
print("Sending message: ", message)
serverSocket.sendto(message, address)

客户端套接字

import socket

UDP_IP = "127.0.0.1"
RPORT = 15000
MESSAGE = "ping"

print("UDP target IP: ", UDP_IP)
print("UDP target port: ", RPORT)
print("message going to send: ", MESSAGE)

client = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)

i=1

while True:
  try:

    if(i<11): 
        client.sendto(MESSAGE.encode('utf_8'),(UDP_IP, RPORT))
        print("sending message: ", MESSAGE)
        print(i)
        i=i+1
        client.settimeout(2)

        data , addr = client.recvfrom(1024)
        print("received echo: ", data)
        print("received at: " , addr )

  finally:
    print("closing socket")
    client.close()

Answer 1

好吧，settimeout() 会产生异常。试试这个:

while True:
  try:
    if(i<11): 
        client.sendto(MESSAGE.encode('utf_8'),(UDP_IP, RPORT))
        print("sending message: ", MESSAGE)
        print(i)
        i=i+1
        client.settimeout(2)

        data , addr = client.recvfrom(1024)
        print("received echo: ", data)
        print("received at: " , addr )

  except socket.timeout:
    print("closing socket")
    client.close()

如果您想了解有关 socket.settimeout() 的更多信息，请查看此链接: http://docs.python.org/2/library/socket.html#socket.timeout

希望对您有所帮助!