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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