Java-如何显示Soap Fault

Question

当Soap Fault调用Java Web Service 时，我会得到这种响应

<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
    <soap:Body>
        <soap:Fault>
            <faultcode>soap:Server</faultcode>
            <faultstring>Fault occurred while processing.</faultstring>
            <detail>
                <ns1:WaybillRegistrationFault xmlns:ns1="http://pod.waybillmanagement.ws.industrysystem.com.ar/">
                    <errors xmlns:ns2="http://pod.waybillmanagement.ws.industrysystem.com.ar/">
                        <code>80000</code>
                        <description>El número de CTG 20140904 ya existe</description>
                    </errors>
                    <errors xmlns:ns2="http://pod.waybillmanagement.ws.industrysystem.com.ar/">
                        <code>1000</code>
                        <description>La carta de porte ya se encuentra registrada.</description>
                    </errors>
                </ns1:WaybillRegistrationFault>
            </detail>
        </soap:Fault>
    </soap:Body>
</soap:Envelope>

我用它的handleFault方法做了一个Soap Handler，如下所示：公共布尔句柄Fault（SOAPMessageContext上下文）｛

    try {
        System.err.println("Handler handleFault");
        if (context.getMessage().getSOAPBody().hasFault()) {
            SOAPFault fa = context.getMessage().getSOAPBody().getFault();
            System.err.println(fa.getFaultString());
            System.err.println("FaultCode: " + fa.getFaultCode() + " - Detail: " + fa.getDetail());
        }
        return true;
    } catch (SOAPException ex) {
        System.err.println("SoapEx " + ex.getMessage());
        return true;
    }
}

但在我的输出中，我只有：

Handler handleFault
Fault occurred while processing.
FaultCode: soap:Server - Detail: [detail: null]

如何处理错误节点？

更新：

使用fa.getDetail().getFirstChild().getTextContent()，我获得xml中的文本。我如何将其作为对象。我认为这是WaybillRegistration的失误。

Answer 1

这是我处理soap错误的方法：

StringWriter sw = new StringWriter();
TransformerFactory.newInstance().newTransformer().transform(
new DOMSource(soapFaultException.getFault()), new StreamResult(sw));
String xml = sw.toString();

Answer 2

如果要将错误内容打印为xml字符串，则必须将fa.getDetail（）返回的dom实例转换为string，fa.getDetails（）.toString（）return["+getNodeName（）+"："+getNode Value（）+"]"而不是xml.的内容

尝试以下操作；

String detailStr=dom2string(fa.getDetail())
System.err.println("FaultCode: " + fa.getFaultCode() + " - Detail: " + detailStr); 

public static final String dom2string(Node node) {
    try {
        if (node == null) {
            return null;
        }
        Transformer tf = transformerThreadLocal.get();
        // Create writer
        StringWriter sw = new StringWriter(buffer, Integer.MAX_VALUE);
        StreamResult result = new StreamResult(sw);
        // transform
        tf.transform(new DOMSource(node), result);
        return sw.getBuffer();
    } catch (Exception e) {
        throw new RuntimeException("Could not convert Node to string", e);
    }
}