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From "Scott Nichol" <snic...@computer.org>
Subject Re: Problem accessing a web service. Please help. Thanks
Date Fri, 11 Jan 2002 22:05:11 GMT
When you create your parameter, you must specify the class of the parameter,
not the Parameter class.  For example,

    String symbol;
    Call call = new Call ();
    // set call properties here
    Vector params = new Vector ();
    // Here it is: String parameter specified as String.class type
    params.addElement (new Parameter("symbol", String.class, symbol, null));
    call.setParams (params);

Scott

----- Original Message -----
From: "Max Stolyarov" <MStolyarov@Novarra.com>
To: <soap-dev@xml.apache.org>
Sent: Friday, January 11, 2002 4:41 PM
Subject: Problem accessing a web service. Please help. Thanks


> Hello,
>
>    I wrote a very sime Web Service and a client application to access. I
> successfully deployed the service on Tomcat, but when I try to make an RPC
> call to it from the client application I get the following error. Can
> somebody please shed some light on it for me. Thanks
>
> SOAP Fault Msg I receive:
>
> SOAPException = SOAP-ENV:Client, No Serializer found to serialize a
> 'org.apache.soap.rpc.Parameter' using encoding style 'SOAP-ENC'.
>
> Max Stolyarov
> NOVARRA
> 3232 Kennicott Ave
> Arlington Heights, Illinois  60004
> Phone:  (847) 368-7800 x 252
> Facsimile:  (847) 590-8144
>


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