"must be str, not bytes" while trying to authenticate to xmpp server

Question

My code is:

import xmpp

jabberid = "[email protected]"
password = "password"
receiver = "[email protected]"
message  = "hello world"

jid = xmpp.protocol.JID(jabberid)
connection = xmpp.Client(server=jid.getDomain())
connection.connect()
connection.auth(user=jid.getNode(), password=password,resource=jid.getResource())

I'm recieving this error:

 File "/home/eduardo/.local/share/virtualenvs/xmpp-sv-KIKvfzDF/lib/python3.6/site-packages/flask/app.py", line 1950, in full_dispatch_request
    rv = self.dispatch_request()
  File "/home/eduardo/.local/share/virtualenvs/xmpp-sv-KIKvfzDF/lib/python3.6/site-packages/flask/app.py", line 1936, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File "/home/eduardo/Desktop/adiconf/xmpp-sv/api/main/user/routes.py", line 33, in add
    connection.auth(user=str(jid.getNode()), password=str(password), resource=str(jid.getResource()))
  File "/home/eduardo/.local/share/virtualenvs/xmpp-sv-KIKvfzDF/lib/python3.6/site-packages/xmpp/client.py", line 228, in auth
    while self.SASL.startsasl=='in-process' and self.Process(1): pass
  File "/home/eduardo/.local/share/virtualenvs/xmpp-sv-KIKvfzDF/lib/python3.6/site-packages/xmpp/dispatcher.py", line 126, in Process
    raise _pendingException[0](_pendingException[1]).with_traceback(_pendingException[2])
  File "/home/eduardo/.local/share/virtualenvs/xmpp-sv-KIKvfzDF/lib/python3.6/site-packages/xmpp/dispatcher.py", line 304, in dispatch
    handler['func'](session,stanza)
  File "/home/eduardo/.local/share/virtualenvs/xmpp-sv-KIKvfzDF/lib/python3.6/site-packages/xmpp/auth.py", line 179, in SASLHandler
    self.DEBUG('Got challenge:'+data,'ok')
TypeError: must be str, not bytes

I'm using the xmpppy module, but I'm not sure why? as the params that I'm passing to connection.auth are all strings

Answer 1

You must decode data in 'auth' file in lib. This line must be looked like that.

data=base64.b64decode(incoming_data).decode()