You need to map the same qualified name as the exception shows:

smr.mapTypes(Constants.NS_URI_SOAP_ENC, new QName("uri:Login:SOAPSDK1", "Answer"), 
null, null, sd);

Scott Nichol

Do not send e-mail directly to this e-mail address,
because it is filtered to accept only mail from
specific mail lists.
----- Original Message ----- 
From: "arvind" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, July 05, 2004 1:25 AM
Subject: msg=No Deserializer found



Hi,
  I have a service installed on MS Soap which calls a VB Dll. I tried to access it 
using Apache Soap. Then I got the following error.

[SOAPException: faultCode=SOAP-ENV:Client; msg=No Deserializer found to deserialize a 
'uri:Login:SOAPSDK1:Answer' using encoding style 'null'.; 
targetException=java.lang.IllegalArgumentException: No Deserializer found to 
deserialize a 'uri:Login:SOAPSDK1:Answer' using encoding style 'null'.]

I have included the following code to Deserialize the incoming string. 

SOAPMappingRegistry smr = new SOAPMappingRegistry ();
    StringDeserializer sd = new StringDeserializer ();
    smr.mapTypes (Constants.NS_URI_SOAP_ENC, new QName ("", "Answer"), null, null, sd);

    // create the transport and set parameters
   SOAPHTTPConnection st = new SOAPHTTPConnection();

   // build the call.
   Call call = new Call ();
   call.setSOAPTransport(st);
   call.setSOAPMappingRegistry (smr);

Please help!!!

Thank you,
Arvind. K

ARVIND

Reply via email to