Python socket.send() can only send once, then socket.error: [Errno 32] Broken pipe occurred

问题

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

Here's the code I wrote:

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()

I ran them in 2 separate terminals:

output of 1st terminal:

$ python2.7 testServer.py
server start
0
1
2
3
4
c1:0
5
6
7
8
9
10
11
...

output of 2nd terminal:

$ 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

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)

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

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

回答1:

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

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).

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.

来源：https://stackoverflow.com/questions/6237569/python-socket-send-can-only-send-once-then-socket-error-errno-32-broken-pi