even in the called function should be enough. On Thursday, June 11, 2015 at 9:39:32 PM UTC+2, Pengfei Yu wrote: > > Yeah, you are right. Stupid question. I am not familiar with how response > works. > > def call(): > response.namespace = 'http://tempuri.org/' > return service() > > > This is enough in call() function. > > Thanks! > > On Thursday, June 11, 2015 at 3:18:13 PM UTC-4, Niphlod wrote: >> >> isn't a simple >> >> def what_you_serve(): >> ......foo bar >> response.namespace = 'http://tempuri.org/' >> return dict(a=1) >> >> working ? >> >> On Thursday, June 11, 2015 at 6:36:31 PM UTC+2, Pengfei Yu wrote: >>> >>> Hi, >>> >>> I want to change the target namespace to "http://tempuri.org/" in the >>> SOAP WSDL xml file. >>> >>> From the pysimplesoap's server.py, I see that I can specify it in >>> SoapDispatcher >>> class with namespace parameter. But I wonder how can I specify it in >>> web2py controller's "@service.soap" decorator. >>> >>> From the code of "/gluon/tools.py", it seems that the namespace could be >>> given by setting "response.namespace" >>> >>> def serve_soap(self, version="1.1"): >>> try: >>> from gluon.contrib.pysimplesoap.server import SoapDispatcher >>> except: >>> return "pysimplesoap not installed in contrib" >>> request = current.request >>> response = current.response >>> procedures = self.soap_procedures >>> >>> >>> location = "%s://%s%s" % ( >>> request.env.wsgi_url_scheme, >>> request.env.http_host, >>> URL(r=request, f="call/soap", vars={})) >>> *namespace = 'namespace' in response and response.namespace or >>> location* >>> documentation = response.description or '' >>> dispatcher = SoapDispatcher( >>> name=response.title, >>> location=location, >>> action=location, # SOAPAction >>> namespace=namespace, >>> prefix='pys', >>> documentation=documentation, >>> ns=True) >>> >>> I wonder how I can set response.namespace directly from web2py >>> controller. A working example will be very nice. >>> >>> Thanks! >>> >>> >>> >>> >>
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