Python socket.send() 只能发送一次，然后socket.error: [Errno 32] Broken pipeline gone

Question

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:) 我使用 Python2.7 在 Ubuntu 10.04.2 上创建了 1 个客户端和 1 个 SocketServer.ThreadingMixIn 服务器，但是 似乎我只能在客户端中调用 sock.send() 一次，然后我会得到：

Traceback (most recent call last):
  File "testClient1.py", line 33, in <module>
    sock.send('c1:{0}'.format(n))   
socket.error: [Errno 32] Broken pipe

这是我编写的代码：

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 2 个单独的终端中运行它们：

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

我尝试在 testClient.py 中直接调用 sock.send() 两次， 例如：

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

但终端的输出仍然相同:( 谁能指出我在这里做错了什么？ 感谢广告！

这是我想出的[Sol]。谢谢马克:)

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()

Answer 1

对于每个连接，在SocketServer.StreamRequestHandler 中调用一次handle()。如果您从 handle 返回，则连接将关闭。

如果您希望服务器处理多个发送/接收，则必须循环直到recv()返回0，这表明客户端关闭了连接（或者至少在发送时调用了shutdown()）。

另请注意，TCP 是一种流协议。您需要设计一个消息协议来指示消息的长度或结尾，并缓冲 recv 直到获得完整的消息。检查 send 返回值以确保所有消息也已发送，或使用 sendall。