2

I'm trying to develop a SOAP Web Service in Django 1.5.2 with Python 2.7.3 and soaplib 0.8.1. Everything works fine at the moment, but now i need to add namespace to @soapmethod response.

This is my view:

from sms.soaplib_handler import DjangoSoapApp, soapmethod, soap_types
from django.views.decorators.csrf import csrf_exempt


class SmsGatewayService(DjangoSoapApp):

    __tns__ = 'http://mismsgw01.milano.ea.it/soap'

    @soapmethod(
        soap_types.String, 
        soap_types.String, 
        soap_types.String, 
        soap_types.Integer,
        soap_types.Boolean, 
        soap_types.Boolean, 
        soap_types.String, 
        _returns=soap_types.Any
    )
    def sendSms(
        self, 
        sendTo, 
        numSender,
        senderDescription,
        timeToLive,
        isDelivered,
        isStatistics,
        messageText
    ):

        retCode = '<retCode>OK</retCode>'

        return retCode

sms_gateway_service = csrf_exempt(SmsGatewayService())

and this is the response when i call the method:

<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/">
   <SOAP-ENV:Body>
      <sendSmsResponse>
         <sendSmsResult>
            <retCode>OK</retCode>
         </sendSmsResult>
      </sendSmsResponse>
   </SOAP-ENV:Body>
</SOAP-ENV:Envelope>

Anyone can help me? How i can modify a soap response? I need to have something similar to:

<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/">
   <SOAP-ENV:Body>
      <sendSmsResponse xmlns="http://mismsgw01.milano.ea.it/soap">
         <retCode>OK</retCode>
      </sendSmsResponse>
   </SOAP-ENV:Body>
</SOAP-ENV:Envelope>

Thanks in advance

kl4us
  • 21
  • 3

1 Answers1

0

You should use Spyne instead of SoapLib. As Spyne was built from soapLib, the spirit of the code will remain and Spyne works fine with its own wsdl.

Julien Greard
  • 969
  • 1
  • 12
  • 32