Does Asyncore Send require newline?

John W jmw136 at gmail.com
Tue Nov 1 04:11:58 CET 2005


Hi Steve,

Here is what I have for my server (This code is really not that pretty, I
apologize):

import socket
import SocketServer

class CommonServer(SocketServer.ThreadingTCPServer):
def server_bind(self):
"""Override server_bind to store the server name."""
SocketServer.TCPServer.server_bind(self)
host, port = self.socket.getsockname()
if not host or host == '0.0.0.0 <http://0.0.0.0>':
host = socket.gethostname()
hostname, hostnames, hostaddrs = socket.gethostbyaddr(host)
if '.' not in hostname:
for host in hostnames:
if '.' in host:
hostname = host
break
self.server_name = hostname
self.server_port = port

print self.server_name
print port

class CommonRequestHandler(SocketServer.StreamRequestHandler):
def handle(self):
print "Handling request"

self.wfile.write('000 Connected ')
datastart = 0
while 1:
print "looping"
s = self.rfile.readline()
print "Message:" + s
if s[-2:] == ' ': s = s[:-2]
elif s[-1:] == ' ': s = s[:-1]

s = s[0:4]
if s == 'QUIT':
print "Quit"
self.wfile.write('000 Bye\n')
break
elif s == 'NAME':
print "Name"
self.wfile.write('000 My name is Python\n')
elif s == 'DATA':
print "Data"
self.wfile.write('000 Data Message\n' )
else:
print "Unknown"
self.wfile.write('200 Unknown Command\n')

def setup(self):
SocketServer.StreamRequestHandler.setup(self)

def test():
portnum = 8001
comserver = CommonServer(('', portnum), CommonRequestHandler )
comserver.handle_request()

print "Running"

if __name__ == '__main__':
test()

Here is what I have for my client:
ASYNC_POLL_LENGTH = 0.05

class Connection(asyncore.dispatcher):
def __init__ (self, server, socket_num):
self.server_name = str( server )
self.port_num = socket_num
self.message_queue = []
self.current_message = 0

self.rcv_data = ""

asyncore.dispatcher.__init__(self)
self.create_socket( socket.AF_INET, socket.SOCK_STREAM )
self.connect(( self.server_name, self.port_num ))

def handle_connect( self ):
print "CONNNECTED"

# collect some more finger server output.
def handle_read_event( self ):
print "handle_read_event received"

self.rcv_data = ""

more = self.recv(512)
print "More Data: %s" % more
self.rcv_data = self.rcv_data + more
print "RCV Data: %s" % self.rcv_data

if self.current_message:
print "Message: %s Type: %s RCV Data: %s" % (self.current_message['data'],
self.current_message['type'], self.rcv_data )
self.current_message['callback']( self.current_message, self.rcv_data )


# the other side closed, we're done.
def handle_close( self ):
print 'Handling Close Event'
self.connected_callback( False )

def handle_write( self ):
# If there is a message in the queue
if len( self.message_queue ) > 0:
# Pop the first message off the queue
self.current_message = self.message_queue.pop(0)
print "Sending Message: %s" % self.current_message['data']
self.send(self.current_message['data'] + "\n")

def queue( self, message ):
self.message_queue.append( message )
print "Message Queued: %s" % message['data']

#========================================================================================
class TestAppClient:
def __init__( self, server, socket ):
self.server = server
self.socket = socket

self.nomad = Connection( server, socket )
asyncore.loop(count=1)

Then I use a timer to call "asyncore.loop(count=1)" every 10ms or on the
IDLE loop of my GUI.

The line in red is what I have to append the '\n' to in order for my server
to get it.

Thanks in advance,

John


On 10/31/05, Steve Holden <steve at holdenweb.com> wrote:
>
> John W wrote:
> > Hello,
> >
> > I have a test environment that consists of a server and an application
> that
> > opens an asyncore socket and communicates with it.
> >
> > In my application, on the handle_write call, if I do not append a "\n"
> to my
> > message the server never gets the message. If I do append it, the server
> can
> > read the message and act accordingly.
> >
> > So my call looks like this:
> > self.send( message] + "\n")
> >
> > Is this how it was designed? Do you have to send a '\n' to have the data
> > read? I can't find a whole lot of documentation on the send method.
> >
> > If you could offer some help, I would appreciate it.
> >
> The short answer to your question is "no, send() does not require
> newline". But much depends on your protocol design and on explicit
> details of your client and server software. You'd probably have to post
> a bit more code before we could say why you are seeing what you are
> seeing.
>
> regards
> Steve
> --
> Steve Holden +44 150 684 7255 +1 800 494 3119
> Holden Web LLC www.holdenweb.com <http://www.holdenweb.com>
> PyCon TX 2006 www.python.org/pycon/ <http://www.python.org/pycon/>
>
>
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