java用axis调用c#的webservice的问题

Question

package test;import javax.xml.namespace.QName;import org.apache.axis.client.Call;
import org.apache.axis.client.Service;public class Test {  /** 
 * @param args 
 */ 
public static void main(String[] args) { 
Object[] orderinfo=new Object[2]; 
orderinfo[0]="北京";
orderinfo[1]=2;
String str = ""; 
try { 
String url = "http://www.ayandy.com/Service.asmx"; 
Service service = new Service(); 
Call call = (Call) service.createCall(); 
call.setTargetEndpointAddress(new java.net.URL(url)); 
call.setOperationName(new QName("http://tempuri.org/", "getWeatherbyCityName"));
call.addParameter(new QName("http://tempuri.org/", "theCityName"),org.apache.axis.encoding.XMLType.XSD_STRING,javax.xml.rpc.ParameterMode.IN);
call.addParameter(new QName("http://tempuri.org/", "theDayFlag"),org.apache.axis.encoding.XMLType.XSD_INTEGER,javax.xml.rpc.ParameterMode.IN);
call.setUseSOAPAction(true); 
call.setSOAPActionURI("http://tempuri.org/getWeatherbyCityName"); 
call.invoke(new Object[]{"常州","Today"});  } catch (Exception e) { 
System.out.println("asdf"); 
e.printStackTrace(); 
} 
System.out.println(str); 
} 

}程序执行后会有异常如下：
faultCode: {http://schemas.xmlsoap.org/soap/envelope/}Server.userException
 faultSubcode: 
 faultString: org.xml.sax.SAXException: SimpleDeserializer encountered a child element, which is NOT expected, in something it was trying to deserialize.请各位高手帮我一下，谢谢

Answer 1

Try this. Thanks.
call.setReturnType(org.apache.axis.encoding.XMLType.XSD_SCHEMA);
Schema result= (Schema)call.invoke(new Object[]{"常州","Today"}); 
for (MessageElement messageElement : result.get_any()) {
System.out.println(messageElement.getAsString());
}