CODEWITHSUNDEEP
pythonxmppbots
Are
Are

Reputation: 179

XMPP connecting to server (Python)

I am currently following a tutorial/getting started on creating a Google Talk Bot

I have made the EchoBot but it fails at runtime with the following error:

File "echobot.py", line 59, in <module>
if xmpp.connect(('talk.google.com'), '5222'):
File "C:\Python31\Lib\sleekxmpp\clientxmpp.py", line 143, in connect
reattempt=reattempt)
File "C:\Python31\Lib\sleekxmpp\xmlstream\xmlstream.py", line 372, in connect
self.address = (host, int(port))
ValueError: invalid literal for int() with base 10: 'a'

I have tried Python 3,3 , 3,2 and 3,1 interpreter with the same result.

I can not get these lines from the tutorial to compile with either of the compilers:

if sys.version_info < (3, 0):
reload(sys)
sys.setdefaultencoding('utf8')

The code from within the Bot where it fails is as follows:

if xmpp.connect(('talk.google.com'), '5222'):
    xmpp.process(block=True)
else:
    print('Fekk ikkje kopla til')

Thanks for any tips/help!

Upvotes: 0

Views: 1315

Answers (1)

User
User

Reputation: 14863

You have written something wrong:

xmpp.connect(('talk.google.com', '5222'))
                               |
                             there you missed a comma

and the second thing could be to put the port to the second place.

xmpp.connect(('talk.google.com', 5222), '5222')

I do not know of the second argument but

Internet Adresses are almost always tuples of (HOST, PORT)

Upvotes: 2

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