All,

I am trying to connect to MSSOAP 3.0 from the Apache Soap shipped with IBM WebSphere Application Server 4.0 (not sure of version), and am getting the following error:

Server:XML Parser failed at linenumber 5, lineposition 36, reason is: Only a default namespace can have an empty URI.

The request that was sent:

POST http://XXXXX/XXXX/XXXXXXXXX.wsdl HTTP/1.0
Host: XXXXX
Content-Type: text/xml; charset=utf-8
Content-Length: 699
SOAPAction: "http://xxxxxx/xxxxxx/xxxxxx/xxxxxx.Login"

<?xml version='1.0' encoding='UTF-8'?>
<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<SOAP-ENV:Body>
<ns1:Login xmlns:ns1="http://tempuri.org/test2/message/" SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/">
<strLoginName xmlns:ns2="" xsi:type="ns2:Result">xxxxxxx</strLoginName>
<strPassword xmlns:ns3="" xsi:type="ns3:Result">xxxxxxxxx</strPassword>
</ns1:Login>
</SOAP-ENV:Body>
</SOAP-ENV:Envelope>

It does not like xmlns:ns2="" -- how do I specify a value for the namespace?  The code creating the call is generated by WSAD and is:

    Vector params = new Vector();
    Parameter strLoginNameParam = new Parameter("strLoginName", java.lang.String.class, strLoginName, Constants.NS_URI_SOAP_ENC);
    params.addElement(strLoginNameParam);
    Parameter strPasswordParam = new Parameter("strPassword", java.lang.String.class, strPassword, Constants.NS_URI_SOAP_ENC);
    params.addElement(strPasswordParam);
    call.setParams(params);
   
    Response resp = call.invoke(getURL(), SOAPActionURI);

Thanks.

Michael Fork
[EMAIL PROTECTED]

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