spring&web服务客户端 – 故障详细信息

我如何获得SoapFaultClientException发送的故障详细信息？
我使用WebServiceTemplate,如下所示：

WebServiceTemplate ws = new WebServiceTemplate();
ws.setMarshaller(client.getMarshaller());
ws.setUnmarshaller(client.getUnMarshaller());
try {
    MyResponse resp = (MyResponse) = ws.marshalSendAndReceive(WS_URI,req);
} catch (SoapFaultClientException e) {
     SoapFault fault =  e.getSoapFault();
     SoapFaultDetail details = e.getSoapFault().getFaultDetail();
      //details always NULL ? Bug?
}

发送的Web服务错误似乎是正确的：

谢谢

Willome

最佳答案

我还遇到了getFaultDetail()返回null(对于SharePoint Web服务)的问题.我可以通过使用类似于此的方法来获取detail元素：

private Element getDetail(SoapFaultClientException e) throws TransformerException {
    TransformerFactory transformerFactory = TransformerFactory.newInstance();
    Transformer transformer = transformerFactory.newTransformer();
    DOMResult result = new DOMResult();
    transformer.transform(e.getSoapFault().getSource(),result);
    NodeList nl = ((Document)result.getNode()).getElementsByTagName("detail");
    return (Element)nl.item(0);
}

之后,您可以在返回的Element或任何您想要的内容上调用getTextContent().