python udp客户端超时机器
python udp客户端超时机器
提问于 2013-08-19 18:16:38
如果服务器套接字中生成的rand数小于4，我的客户套接字将在接收数据时挂起。我需要设置超时机制，允许客户套接字检测到有“超时”，然后它将继续发送消息。在运行服务器套接字和客户端套接字之后，显示了以下错误消息：
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服务器插座：
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()回答 1
Stack Overflow用户
发布于 2013-12-20 17:53:10
那么，settimeout()会产生一个异常。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
希望这能有所帮助！
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/18311338
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