解析肥皂wsdl java

Question

我无法解析soap请求wsdl文件。我有来自soapui的xml文件 （soapui-project.xml）和此文件中的数据id和名称

<types>
 <xsd:schema>
  .......
 </xsd:schema>
 </type>....
...<soapenv:Header/>
<soapenv:Body>
<!-Optional:-->
<idNumber>123</idNumber>
<name>abc</name>...

我尝试使用java代码：

 public SOAPMessage readSoapMessage(String xmlInput) throw SOAPException, FileNotFoundException{
         SOAPMessage message = MessageFactory.newInstance().createMessage();
         SOAPPart soapPart = message.getSOAPPart();
          soapPart.setContent(new StreamSource(new FileInputStream(xmlInput)));
          message.saveChanges();
          return message;
}
SOAPMessage message = readSoapMessage(xmlInput);
message.writeTo(System.out);

但我得到的所有xml文件我只需要获取idNumber和name。我怎么能这样做？ 谢谢