From jc at eclis.ch  Wed Mar  4 11:01:24 2009
From: jc at eclis.ch (Jean-Christian de Rivaz)
Date: Wed, 04 Mar 2009 11:01:24 +0100
Subject: [PyXMPP] Trying to make a class XMLRPCHandler
Message-ID: <49AE5174.50902@eclis.ch>

Hello,

I try to add a XML-RPC handler (XEP-0009) to the "echobot.py" example. 
So far I have this function and class (edited for brevity):

def test_string_get():
         return ("foo")

class XMLRPCHandler(object):
     implements(IIqHandlersProvider, IFeaturesProvider)
     def __init__(self, client):
         self.client = client
     def get_features(self):
         return ["jabber:iq:rpc"]
     def get_iq_get_handlers(self):
         return []
     def get_iq_set_handlers(self):
         return [("query", "jabber:iq:rpc", self.set_rpc),]
     def set_rpc(self,iq):
         query = iq.get_query().children
         stanza = query.serialize()
         print stanza
         call = xmlrpclib.loads(stanza)
         print call[1], call[0]
         result = locals()[call[1]](*call[0])
         print result
         stanza = "<methodResponse>" + xmlrpclib.dumps((result,)) + 
"</methodResponse>"
         print stanza
         iq=iq.make_result_response()
         q=iq.new_query("jabber:iq:rpc")
         # FIXME: Need to add the methodResponse XML tree.
         # How to do that ?
         return iq

The class has been added into the interface_providers list of the class 
Client. Test show that the XMLRPCHandler.set_rpc() is called as expected.

The problem I have is that the xmlrpclib need a string, not a XML tree. 
Using the serialize() should do the conversion, but the result have the 
"default1:" in every nodes that make it incompatible with the xmlrpclib.

Here is an example fo the serialization result:

<default1:methodCall>
<default1:methodName>test_string_get</default1:methodName>
<default1:params>
</default1:params>
</default1:methodCall>

 From where cam the "default1:" and how to disable it ?

Regards,
-- 
Jean-Christian de Rivaz

From jajcus at jajcus.net  Wed Mar  4 18:38:36 2009
From: jajcus at jajcus.net (Jacek Konieczny)
Date: Wed, 4 Mar 2009 18:38:36 +0100
Subject: [PyXMPP] Trying to make a class XMLRPCHandler
In-Reply-To: <49AE5174.50902@eclis.ch>
References: <49AE5174.50902@eclis.ch>
Message-ID: <20090304173836.GD6146@lolek.nigdzie>

On Wed, Mar 04, 2009 at 11:01:24AM +0100, Jean-Christian de Rivaz wrote:
> The problem I have is that the xmlrpclib need a string, not a XML tree. 
> Using the serialize() should do the conversion, but the result have the 
> "default1:" in every nodes that make it incompatible with the xmlrpclib.
> 
> Here is an example fo the serialization result:
> 
> <default1:methodCall>
> <default1:methodName>test_string_get</default1:methodName>
> <default1:params>
> </default1:params>
> </default1:methodCall>
> 
>  From where cam the "default1:"

It is the default prefix that libxml2 uses for a namespace set for this
object. Namespace declaration is not visible, as it is at the outer
element (<query/>).

> and how to disable it ?

Try using pyxmpp.xmlextra.replace_ns()

Greets,
        Jacek

From jc at eclis.ch  Wed Mar  4 20:49:42 2009
From: jc at eclis.ch (Jean-Christian de Rivaz)
Date: Wed, 04 Mar 2009 20:49:42 +0100
Subject: [PyXMPP] Trying to make a class XMLRPCHandler
In-Reply-To: <20090304173836.GD6146@lolek.nigdzie>
References: <49AE5174.50902@eclis.ch> <20090304173836.GD6146@lolek.nigdzie>
Message-ID: <49AEDB56.4000303@eclis.ch>

Jacek Konieczny a ?crit :
> On Wed, Mar 04, 2009 at 11:01:24AM +0100, Jean-Christian de Rivaz wrote:
>> The problem I have is that the xmlrpclib need a string, not a XML tree. 
>> Using the serialize() should do the conversion, but the result have the 
>> "default1:" in every nodes that make it incompatible with the xmlrpclib.
>>
>> Here is an example fo the serialization result:
>>
>> <default1:methodCall>
>> <default1:methodName>test_string_get</default1:methodName>
>> <default1:params>
>> </default1:params>
>> </default1:methodCall>
>>
>>  From where cam the "default1:"
> 
> It is the default prefix that libxml2 uses for a namespace set for this
> object. Namespace declaration is not visible, as it is at the outer
> element (<query/>).

I am not certain to understand. If I try this simple code below, I don't 
get a "default1:" in any node:

Code:
-----------------
#!/usr/bin/python
import libxml2

text_in = "<foo><bar/></foo>"
print text_in, type(text_in)

tree = libxml2.parseDoc(text_in).children
print tree, type(tree)

text_out = tree.serialize()
print text_out, type(text_out)
-----------------

Result:
<foo><bar/></foo> <type 'str'>
<foo><bar/></foo> <type 'instance'>
<foo><bar/></foo> <type 'str'>

Where is the difference with the jabber.iq.rpc child node parsing and 
serialization ?

I an other point of view: is there a way to simply get the original 
string of the jabber.iq.rpc stanza without doing the parsing (required 
for pyxmpp) and a serialization (because I don't know a other way to 
pass the request to xmlrpclib).

>> and how to disable it ?
> 
> Try using pyxmpp.xmlextra.replace_ns()

Great, PyXMPP have nice resources!
Now a have a working XEP-0009 with this code:

     def set_rpc(self,iq):
         print iq.serialize()
         query = iq.get_query().children
         pyxmpp.xmlextra.replace_ns(query, query.ns(), 
pyxmpp.xmlextra.common_ns)
         stanza = query.serialize()
         print stanza
         call = xmlrpclib.loads(stanza)
         result = globals()[call[1]](*call[0])
         stanza = "<methodResponse>" + xmlrpclib.dumps((result,)) + 
"</methodResponse>"
         print stanza
         tree = libxml2.parseDoc(stanza).children
         iq=iq.make_result_response()
         q=iq.new_query("jabber:iq:rpc")
         q.addChild(tree)
         return iq

Thanks for your help!
-- 
Jean-Christian de Rivaz