如何处理Java中的肥皂错误

Question

我有这样的东西

public void ValidateXML(String attributeName, String columnName) throws JSONException, SQLException {
        NodeList nodeList = responseXML.getElementsByTagName("Profile"); 
        int i=0;
        HashMap<String,String> data = null;
       for (i = 0; i < nodeList.getLength(); i++) {
      Node node = nodeList.item(i);
      if (node.getNodeType() == Node.ELEMENT_NODE) {
        Element element = (Element) node;
        try
        {       
        String pernr =getValue("name", element);  

以上内容是针对SOAP的正确响应

我想知道如何处理以下类似问题

<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
    <soap:Body>
        <soap:Fault xmlns:soap="http://schemas.xmlsoap.org/soap/envelope">
            <faultcode>soap:client</faultcode>
            <faultstring> The value '12345' of element 'abc' is not valid.</faultstring>
        </soap:Fault>
    </soap:Body>
</soap:Envelope>