python does not connect to local XMPP server

Question

i'm trying to connect my local XMPP server by the code coming below

import xmpp
client = xmpp.Client('localhost',debug=[])
client.connect(server=('localhost',5222))

but i always get this message :

An error occurred while looking up _xmpp-client._tcp.localhost

i've checked that the port 5222 is already opened(by using telnet). (i have to mention that the firewall on the localhost is off) now what should i add to this code to make it work ?

Answer 1

This message (a warning, not an error as pointed out in xinox's answer) is indicating that a DNS SRV record lookup failed. DNS SRV records are used to find services that are associated with a certain domain (eg. localhost in this case, so not really a domain at all which is why the lookup is failing), but which delegate their responsibility to a server living somewhere else.

For instance, if I have a server at example.net , making my Jabber ID (JID): user@example.net , but my XMPP server lived at chat.example.net I could construct an SRV record on example.net to point to chat.example.net . There are other ways to delegate responsibility, but this is the preferred one. XMPPs use of SRV records is defined in RFC 6120 §3.2.1 .

To actually get rid of this error you can use the use_srv kwarg, making your initial example:

import xmpp
client = xmpp.Client('localhost',debug=[])
client.connect(server=('localhost',5222), use_srv=False)

Answer 2

use this.

client = xmpp.Client('127.0.0.1',debug=[])
client.connect(server=('127.0.0.1',5222))

or your IP 192.XXX

Answer 3

i figured this out , just remember that this is a warning not just an error . the python connected to openfire XMPP server correctly and it works fine .