Python socket.send（）只能发送一次，然后出现socket.error：[Errno 32]发生管道损坏

本文介绍了Python socket.send（）只能发送一次，然后出现socket.error：[Errno 32]发生管道损坏的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我是网络编程的新手，所以如果这是一个愚蠢的问题，请原谅我:) 我在Ubuntu 10.04.2上使用Python2.7创建了1个客户端和1个SocketServer.ThreadingMixIn服务器，但是似乎我只能在客户端调用sock.send（）一次，然后得到一个：

I'm a newbie in network programming, so please forgive me if this is a dumb question :) I created 1 client and 1 SocketServer.ThreadingMixIn server on Ubuntu 10.04.2 using Python2.7, but it seems like I can only call sock.send() once in client, then I'll get a:

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：

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：

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()

我在两个单独的终端中运行它们：

I ran them in 2 separate terminals:

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

我尝试直接在testClient.py中两次调用sock.send（），例如：

I tried calling sock.send() twice directly in testClient.py, for ex:

while n<=1000: sock.send('c1:{0}'.format(n)) sock.send('12333') result=sock.recv(1024) print(result) n+=1 time.sleep(1)

但终端的输出仍然相同:( 有人可以指出我在这里做错什么吗？ ！

but the outputs of the terminals are still the same :( Can anyone please point out what am I doing wrong here? Thx in adv!

这是我想出的[Sol]。谢谢Mark：）

Here's the [Sol] I came up with. Thank you Mark:)

testClient1.py：

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：

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()

推荐答案

handle（）在 SocketServer.StreamRequestHandler 中为每个连接调用一次。如果从句柄返回，则连接已关闭。

handle() is called in the SocketServer.StreamRequestHandler once for each connection. If you return from handle the connection is closed.

如果您希望服务器处理多个发送/ recv，您必须循环播放，直到recv（）返回0，表示客户端关闭了连接（或至少在发送时调用了shutdown（））。

If you want the server to handle more than one send/recv, you must loop until recv() returns 0, indicating the client closed the connection (or at least called shutdown() on sends).

还要注意，TCP是流协议。您需要设计一个消息协议，以指示消息的长度或结尾，并缓冲 recv 直到收到完整的消息。检查发送的返回值以确保所有消息也都已发送，或者使用 sendall 。

Also note that TCP is a streaming protocol. You'll need to design a message protocol that indicates the length or end of a message, and buffer recv until you have a complete message. Check send return value to make sure all the message is sent as well, or use sendall.