Python socket.send() 只能发送一次,然后出现 socket.error: [Errno 32] Broken pipe
我刚开始学习网络编程,所以如果这个问题很傻,请多包涵 :) 我在Ubuntu 10.04.2上用Python2.7创建了一个客户端和一个SocketServer.ThreadingMixIn服务器,但似乎我在客户端只能调用一次sock.send(),然后就会出现:
Traceback (most recent call last):
File "testClient1.py", line 33, in <module>
sock.send('c1:{0}'.format(n))
socket.error: [Errno 32] Broken pipe
这是我写的代码:
testClient1.py:
#! /usr/bin/python2.7
# -*- coding: UTF-8 -*-
import sys,socket,time,threading
sock=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
try:
sock.connect(('localhost',20000))
except socket.error:
print('connection error')
sys.exit(0)
n=0
while n<=1000:
sock.send('c1:{0}'.format(n))
result=sock.recv(1024)
print(result)
n+=1
time.sleep(1)
testServer.py:
#! /usr/bin/python2.7
# -*- coding: UTF-8 -*-
import threading,SocketServer,time
class requestHandler(SocketServer.StreamRequestHandler):
#currentUserLogin={} #{clientArr:accountName}
def handle(self):
requestForUpdate=self.rfile.read(4)
print(requestForUpdate)
self.wfile.write('server reply:{0}'.format(requestForUpdate))
class broadcastServer(SocketServer.ThreadingMixIn, SocketServer.TCPServer):
pass
if __name__ == '__main__':
server=broadcastServer(('localhost',20000),requestHandler)
t = threading.Thread(target=server.serve_forever)
t.daemon=True
t.start()
print('server start')
n=0
while n<=60:
print(n)
n+=1
time.sleep(1)
server.socket.close()
我在两个不同的终端运行它们:
第一个终端的输出:
$ python2.7 testServer.py
server start
0
1
2
3
4
c1:0
5
6
7
8
9
10
11
...
第二个终端的输出:
$ python2.7 testClient1.py
server reply:c1:0
Traceback (most recent call last):
File "testClient1.py", line 33, in <module>
sock.send('c1:{0}'.format(n))
socket.error: [Errno 32] Broken pipe
我试着在testClient.py中直接调用sock.send()两次,比如:
while n<=1000:
sock.send('c1:{0}'.format(n))
sock.send('12333')
result=sock.recv(1024)
print(result)
n+=1
time.sleep(1)
但是终端的输出还是一样的 :( 有没有人能告诉我我哪里做错了? 谢谢大家!
这是我想到的[Sol]。谢谢你,Mark :)
testClient1.py:
import sys,socket,time
sock=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
try:
sock.connect(('localhost',20000))
except socket.error:
print('connection error')
sys.exit(0)
n=0
while n<=10: #connect once
sock.send('c1:{0}'.format(n))
result=sock.recv(1024)
print(result)
n+=1
time.sleep(1)
sock.close()
#once you close a socket, you'll need to initialize it again to another socket obj if you want to retransmit
sock=socket.socket(socket.AF_INET,socket.SOCK_STREAM)
try:
sock.connect(('localhost',20000))
except socket.error:
print('connection error')
sys.exit(0)
n=0
while n<=10: #connect once
sock.send('c3:{0}'.format(n))
result=sock.recv(1024)
print(result)
n+=1
time.sleep(1)
sock.close()
testServer.py:
import threading,SocketServer,time
class requestHandler(SocketServer.StreamRequestHandler):
#currentUserLogin={} #{clientArr:accountName}
def handle(self):
requestForUpdate=self.request.recv(1024)
print(self.client_address)
while requestForUpdate!='':
print(requestForUpdate)
self.wfile.write('server reply:{0}'.format(requestForUpdate))
requestForUpdate=self.request.recv(1024)
print('client disconnect')
class broadcastServer(SocketServer.ThreadingMixIn, SocketServer.TCPServer):
pass
if __name__ == '__main__':
server=broadcastServer(('localhost',20000),requestHandler)
t = threading.Thread(target=server.serve_forever)
t.daemon=True
t.start()
print('server start')
n=0
while n<=60:
print(n)
n+=1
time.sleep(1)
server.socket.close()
1 个回答
17
在每次有新的连接时,SocketServer.StreamRequestHandler
会调用一次 handle()
。如果你在 handle
中返回,连接就会关闭。
如果你想让服务器处理多次发送和接收数据,你需要在一个循环中运行,直到 recv()
返回 0,这表示客户端关闭了连接(或者至少在发送时调用了 shutdown())。
另外要注意,TCP 是一种流式协议。你需要设计一种消息协议,用来指示消息的长度或结束,并在接收到完整消息之前缓存 recv
的数据。检查 send
的返回值,以确保所有消息都已发送,或者使用 sendall
来确保发送完整。