Re: [Xmpppy-devel] "After registration" issue

[Yann L. <as...@la...>]

On 01/04/2013 05:05 PM, Gábor László wrote:
> Hi!
>
> My task is to do a Python script, what is registers a new user to a
> jabberd2 server and then assigns all users to this newly created user
> (roster-items). These things works separately (xmpp.features.register,
> ... roster.setItem ... ), but when I do a registration and right after
> that authenticate and set a roster item, that will not happen. Here is
> an example code:
>
>
> import sys
> import os
> import time
> import xmpp
>
> cl = xmpp.Client('jabber.digicpictures.local')
> cl.connect()
>
> xmpp.features.register(cl, 'jabber.digicpictures.local', {'username':
> 'teszt', 'password': 'teszt'})
>
> cl.auth('teszt', 'teszt')
>
> cl.sendInitPresence()
>
> # This will not work, throws an exception (not authorized)
> # If the "xmpp.register" line deleted  (and the user is exists) this
> # will NOT throw an exception
> roster = cl.getRoster()
> roster.setItem('teszt2@jabber.digicpictures.local', 'Teeeszt');
>
> time.sleep(1)
> cl.disconnect()
>
>
> I attached the DEBUG output. The interesting lines are:
>
> DEBUG: dispatcher   wait  Waiting for ID:2 with timeout 25...
> DEBUG: bind         error Binding failed: timeout expired.
>
> Why this is happening? I tried "time.sleep(5)" after register, but nothing.
> Can somebody help me, please? What is the problem or what I do wrong?
> Thank you!

Probably because you don't way the answer before doing next step.

When you do xmpp.features.register() you don't really register, you 
queue a message that will request the registration. message will really 
be sent when you'll call proceede()

-- 
Yann