spring & web service 客户端 - 故障详情

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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 服务故障似乎是正确的:

<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<soapenv:Body>
  <soapenv:Fault>
     <faultcode>soapenv:Client</faultcode>
     <faultstring>Validation error</faultstring>
     <faultactor/>
     <detail>
        <ws:ValidationError xmlns:ws="http://ws.x.y.com">ERR_UNKNOWN</ws:ValidationError>
     </detail>
  </soapenv:Fault>
</soapenv:Body>

谢谢

威洛姆

Answer 1

我还遇到了 getFaultDetail() 返回 null(对于 SharePoint Web 服务)的问题。我可以使用与此类似的方法获取 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，您可以对返回的 Element 或您想要的任何内容调用 getTextContent()。