SleekXMPP send a message at will and still listen for incoming messages.

Question

I am trying to modify the sleekxmpp echobot example to act more like a client. I am able to receive and print messages with the code below but how do I make the code also allow me to send messages to another JID at will and not as an auto response to a message received. I am relatively new to python so any help will be very useful THANK YOU. ``` #!/usr/bin/env python # -*- coding: utf-8 -*- """ SleekXMPP: The Sleek XMPP Library Copyright (C) 2010 Nathanael C. Fritz This file is part of SleekXMPP. See the file LICENSE for copying permission. """ import sys import logging import getpass from optparse import OptionParser import sleekxmpp # Python versions before 3.0 do not use UTF-8 encoding # by default. To ensure that Unicode is handled properly # throughout SleekXMPP, we will set the default encoding # ourselves to UTF-8. if sys.version_info < (3, 0): reload(sys) sys.setdefaultencoding('utf8') else: raw_input = input class EchoBot(sleekxmpp.ClientXMPP): """ A simple SleekXMPP bot that…

Accepted Answer

I have figure out this problem. this problem can use xmlstream._start_thread to fix. but it still a litte ugly.

```
import sys
import logging
import getpass
from optparse import OptionParser
import sleekxmpp

reload(sys)
sys.setdefaultencoding('utf8')

class EchoBot(sleekxmpp.ClientXMPP):
    u"""
    a simple sleekxmmp bot that will echo message
    """

def __init__(self, jid, pwd):
        sleekxmpp.ClientXMPP.__init__(self, jid, pwd)

self.add_event_handler('session_start', self.start)

self.add_event_handler('message', self.message)

def start(self, event):
        u"""
        process session start
        """
        self.send_presence()
        self.get_roster()
        self._start_thread("chat_send", self.chat_send)

def message(self, msg):
        if msg['type'] in ('chat', 'normal'):
            print "others send: %s" % msg
            msg.reply('Thanks for sending
%(body)s' % msg).send()

def chat_send(self):
        while True:
            line = raw_input("send: ")
            if line == "quit":
                break
            print "you send: %s" % line
            self.send_message("test2@dp-pc", line)

if __name__ == "__main__":
    optp = OptionParser()

optp.add_option('-q', '--quiet', help='set logging to error',
                action="store_const", dest="loglevel",
                const=logging.ERROR, default=logging.INFO)
    optp.add_option('-d', '--debug', help='set logging to debug',
                action="store_const", dest="loglevel",
                const=logging.DEBUG, default=logging.INFO)
    optp.add_option('-v', '--verbose', help='set logging to comm',
                action="store_const", dest="loglevel",
                const=5, default=logging.INFO)

optp.add_option('-j', '--jid', dest="jid",
                help="jid to use")
    optp.add_option('-p', '--password', dest='password',
                help="password to use")

opts, args = optp.parse_args()

logging.basicConfig(level=opts.loglevel,
                    format='%(levelname)-8s %(message)s')

if opts.jid is None:
        opts.jid = raw_input("username:")
    if opts.password is None:
        opts.password = getpass.getpass("Password: ")

xmpp = EchoBot(opts.jid, opts.password)
    xmpp.register_plugin('xep_0030')  # service discovery
    xmpp.register_plugin('xep_0004')  # date form
    xmpp.register_plugin('xep_0060')  # pubsub
    xmpp.register_plugin('xep_0199')  # xmpp ping

if xmpp.connect():
        print("xmpp start")
        xmpp.process(block=True)
        print("Done")
    else:
        print("unable to connected")
```

SleekXMPP send a message at will and still listen for incoming messages.

Solution

Problem