Try this...
QName qn = new QName("somenamespace","somequalifiedname");
call.registerTypeMapping(yourbean.class,
qn,
new BeanSerializerFactory(yourbean.class,
qn),
new BeanDeserializerFactory(yourbean.class,
qn));
If you control the service, you'll need to tell the service how to
deserialize. That is done in
the server-config.wsdd block for the service...
<beanMapping languageSpecificType="java:com.yournamesapce.yourbean" qname="
ns1:somequalifiedname" xmlns:ns1="somenamespace"/>
The namespace and qualified name need to match on both ends...
Hope this helps,
Mark Malinoski
Consultant
AES/PHEAA
"Patrick Quinn"
<[EMAIL PROTECTED]
olving.com> To
<[email protected]>
06/24/2005 10:08 cc
AM
Subject
deserializing error
Please respond to
[EMAIL PROTECTED]
he.org
Hi
Has anyone seen an error like this before, or know what needs to be done to
cure it?
org.xml.sax.SAXException: Deserializing parameter 'ProvidentResponse':
could not find deserializer for type
{http://ProvidentConnector.ProvidentResponseToOrch}ProvidentResponse
The error occurs when I make the following call to the service:
String result =
pt.SOPResponse(prvResp);
where prvResp is a bean comprised of four String fields.
My WSDL would appear to be fine, so I don't think that's the problem.
Am I missing code for the deserializer, or am I missing something from the
CLASSPATH (although I would expect a different error were this the case)?
Thanks in advance
Pat
______________________________________________________________________
This email has been scanned by the MessageLabs Email Security System.
For more information please visit http://www.messagelabs.com/email
______________________________________________________________________
