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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