calling a webservice without arguments

Asked by oyzzo on 2012-11-15

Hi, I'm trying to call a webservice which has no input arguments and returns a list of custom type elements .
I call it like this:

             client.getAllProjectes({}, null,
                        function(resp) {
                                                document.getElementById("demo").innerHTML=resp;
                        }
                )

I've also tryied null instead of {}.
This is what i'm getting as response:

{"fault": {"lineno": 166, "filename": "dispatcher", "code": "server", "string": "too many values to unpack", "detail": ["Traceback (most recent call last):", " File \"/usr/local/lib/python2.7/dist-packages/ladon-0.7.5-py2.7.egg/ladon/server/dispatcher.py\", line 159, in dispatch_request", " result = self.call_method(method,req_dict,tc,export_dict,log_line)", " File \"/usr/local/lib/python2.7/dist-packages/ladon-0.7.5-py2.7.egg/ladon/server/dispatcher.py\", line 96, in call_method", " result = getattr(service_class_instance,req_dict['methodname'])(*args)", " File \"/usr/local/lib/python2.7/dist-packages/ladon-0.7.5-py2.7.egg/ladon/ladonizer/decorator.py\", line 77, in injector", " res = f(*args,**kw)", " File \"/var/www/wsgi/gesalight/projecte.py\", line 91, in getAllProjectes", " for idx,tipus,codi,titol,obs,cli in projs:", "ValueError: too many values to unpack", ""]}, "version": "1.0", "type": "jsonwsp/fault"}

What am i doing wrong?

Question information

Language:
English Edit question
Status:
Solved
For:
ladon Edit question
Assignee:
No assignee Edit question
Solved by:
oyzzo
Solved:
2012-11-15
Last query:
2012-11-15
Last reply:
oyzzo (oyzzoforfree) said : #1

Sorry it's a problem on the python server side.