如何在 spring-ws 中解析 SoapFaultClientException

Question

我正在使用spring-ws-2.3.1，在为Web服务创建客户端时，有时我得到SoapFaultClientException，如下所示，

<SOAP-ENV:Body>
  <SOAP-ENV:Fault>
     <faultcode>SOAP-ENV:Server</faultcode>
     <faultstring>There was a problem with the server so the message could not proceed</faultstring>
     <faultactor>InvalidAPI</faultactor>
     <detail>
        <ns0:serviceException xmlns:ns0="http://www.books.com/interface/v1">
           <ns1:messageId xmlns:ns1="http://www.books.org/schema/common/v3_1">5411</ns1:messageId>
           <ns1:text xmlns:ns1="http://www.books.org/schema/common/v3_1">Locale is invalid.</ns1:text>
        </ns0:serviceException>
     </detail>
  </SOAP-ENV:Fault>

我试图获取服务异常的"消息 ID"和"文本"，但我不能。请在下面找到代码，

catch (SoapFaultClientException ex) {
        SoapFaultDetail soapFaultDetail = ex.getSoapFault().getFaultDetail(); // <soapFaultDetail> node
        // if there is no fault soapFaultDetail ...
        if (soapFaultDetail == null) {
            throw ex;
        }
        SoapFaultDetailElement detailElementChild = soapFaultDetail.getDetailEntries().next();
        Source detailSource = detailElementChild.getSource();
        Object detail = webServiceTemplate.getUnmarshaller().unmarshal(detailSource);
        System.out.println("Detail"+detail.toString());//This object prints the jaxb element
    }

"detail"对象返回 JaxbElement.Is 有任何优雅的方式来解析肥皂错误。

任何帮助都应不胜感激。

Answer 1

最后，我可以解析肥皂错误异常，

catch (SoapFaultClientException ex) {
    SoapFaultDetail soapFaultDetail = ex.getSoapFault().getFaultDetail(); // <soapFaultDetail> node
    // if there is no fault soapFaultDetail ...
    if (soapFaultDetail == null) {
        throw ex;
    }
    SoapFaultDetailElement detailElementChild = soapFaultDetail.getDetailEntries().next();
    Source detailSource = detailElementChild.getSource();
    Object detail = webServiceTemplate.getUnmarshaller().unmarshal(detailSource);
    JAXBElement<serviceException> source = (JAXBElement<serviceException>)detail;
    System.out.println("Text::"+source.getText()); //prints : Locale is invalid.
}

我找不到任何其他优雅的方式，所以我希望这应该是解决方案。

Answer 2

这样做，没有解组。

            String DETAIL_PATH = "//*[local-name()='Detail']";
            SoapFaultDetail soapFaultDetail = ((SoapFaultClientException)exception).getSoapFault().getFaultDetail();
            DOMSource sourceNode = (DOMSource) soapFaultDetail.getSource();
            XPath xPath = XPathFactory.newInstance().newXPath();
            XPathExpression exp = null;
            //Cibling detail field
            try {
                exp = xPath.compile(DETAIL_PATH);
                NodeList nl = (NodeList) exp.evaluate(sourceNode.getNode(), XPathConstants.NODESET);
                if (nl == null || nl.getLength() < 1) {
                    return;
                } else {
                    //Get value here
                    String detailValue = nl.item(0).getTextContent();
                }
            }
            catch (XPathExpressionException e) {
                LOG.debug("Can't get detail from Exception");
            }