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