Hi Patrick,
you have a generated stub class. look at your stub class if this code
segement exists
qName = new javax.xml.namespace.QName("urn:[yourpacckagename]"
"ProvidentResponse");
cachedSerQNames.add(qname);
cls = yourpackagename.ProvidentResponse.class;
cachedSerClasses.add(cls);
cachedSerFactories.add(beansf);
cachedDeserFactories.add(beandf);
This is the code which must be in your stub class. Also you must be sure
that you have deployed the service with the right typeMapping how it's
described by Mark.
Regards
Ferruh
> --- Ursprüngliche Nachricht ---
> Von: "Patrick Quinn" <[EMAIL PROTECTED]>
> An: <axis-user@ws.apache.org>
> Betreff: RE: deserializing error
> Datum: Fri, 24 Jun 2005 17:00:32 +0100
>
> Thanks for the tip.
>
> The registerTypeMapping call is already made in the stub, as part of
> createCall():
>
>
> synchronized (this) {
> if (firstCall()) {
> // must set encoding style before registering
> serializers
> _call.setEncodingStyle(null);
> for (int i = 0; i < cachedSerFactories.size(); ++i)
> {
> java.lang.Class cls = (java.lang.Class)
> cachedSerClasses.get(i);
> javax.xml.namespace.QName qName =
> (javax.xml.namespace.QName)
> cachedSerQNames.get(i);
> java.lang.Class sf = (java.lang.Class)
> cachedSerFactories.get(i);
> java.lang.Class df = (java.lang.Class)
> cachedDeserFactories.get(i);
> _call.registerTypeMapping(cls, qName, sf, df,
> false);
> }
> }
> }
> return _call;
>
> Should I add your code segments to my client implementation despite
> this?
>
> Thanks
>
> Pat
>
> -----Original Message-----
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> Sent: 24 June 2005 16:43
> To: axis-user@ws.apache.org
> Subject: Re: deserializing error
>
>
> 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
> <axis-user@ws.apache.org>
>
> 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
>
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