jshiffer/slixmpp
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

wait, shouldn't 'port' default to 5222?? Would seem logical to me.

Browse Source
This commit is contained in:
Tom Nichols 2010-07-09 16:15:18 -04:00
parent 9506970042
commit 969c4652a4
1 changed files with 3 additions and 3 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
sleekxmpp/xmlstream/xmlstream.py
View File

@ -52,7 +52,7 @@ DEFAULT_KEEPALIVE = 300 # send a single byte every 5 minutes
class XMLStream(object):
"A connection manager with XML events."
def __init__(self, socket=None, host='', port=0, escape_quotes=False):
def __init__(self, socket=None, host='', port=5222, escape_quotes=False):
global ssl_support
self.ssl_support = ssl_support
self.escape_quotes = escape_quotes
@ -101,7 +101,7 @@ class XMLStream(object):
def setFileSocket(self, filesocket):
self.filesocket = filesocket
def connect(self, host='', port=0, use_ssl=None):
def connect(self, host='', port=5222, use_ssl=None):
"Establish a socket connection to the given XMPP server."
if not self.state.transition('disconnected','connected',
@ -117,7 +117,7 @@ class XMLStream(object):
# TODO currently a caller can't distinguish between "connection failed" and
# "we're already trying to connect from another thread"
def connectTCP(self, host='', port=0, use_ssl=None, reattempt=True):
def connectTCP(self, host='', port=5222, use_ssl=None, reattempt=True):
"Connect and create socket"
# Note that this is thread-safe by merit of being called solely from connect() which