Hi Alexandre,
I've done something like this in my controller:
def __call__(self, environ, start_response):
# Wrap an exception in an XMLRPC fault
try:
return XMLRPCController.__call__(self, environ, start_response)
except:
logException()
if asbool(config['global_conf']['debug']):
return xmlrpc_fault(0,
traceback.format_exc())(environ, start_response)
else:
return xmlrpc_fault(0, "An internal server error
occurred")(environ, start_response)
finally:
Session.remove()
Note the call to the return value of xmlrpc_fault with (environ, start_response)
HTH,
Chris
On Feb 4, 2008 5:53 AM, Alexandre Conrad <[EMAIL PROTECTED]> wrote:
>
> I'd like to put this post back up, hoping for a reply...
>
>
>
> Alexandre Conrad wrote:
>
> > Hi,
> >
> > I would like to know how errors should be handled having webservices
> > running with XMLRPCController.
> >
> > Say I have a "def login(self, user, password)" method. If the login
> > fails, how should I return the errors to the client? I saw, from the
> > following docs:
> >
> > http://wiki.pylonshq.com/display/pylonsdocs/Using+the+XMLRPCController
> >
> > that a convenient fault handler function is provided:
> >
> > def xmlrpc_fault(code, message):
> > """Convenience method to return a Pylons response XMLRPC Fault"""
> >
> > So I imported this function in my controller:
> >
> > from pylons.controllers.xmlrpc import xmlrpc_fault
> >
> > and added to my login method:
> >
> > user = model.User.query.filter_by(user=user, password=password).one()
> > if not user:
> > return xmlrpc_fault(LOGIN_ERR, "incorrect login")
> >
> > Is this how it should be handled ? If it is, I have the following error
> > message:
> >
> > exceptions.TypeError: cannot marshal <class
> > 'paste.wsgiwrappers.WSGIResponse'> objects
> >
> > Regards,
>
> --
> Alexandre CONRAD
>
>
> >
>
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