解析绿洲XML文档

Question

我想找到使用OASIS XSD解析响应的最佳方法。我有以下响应XML字符串，但不确定如何解析它，我尝试从OASIS复制XSD，但未创建JAXB对象。这是XML响应数据：

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
<soap:Header>
    <wsse:Security xmlns:wsse="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd">
    <wsu:Timestamp xmlns:wsu="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-utility-1.0.xsd" wsu:Id="Timestamp-521833581">
    <wsu:Created>2017-09-07T00:00:02.002Z</wsu:Created>
</wsu:Timestamp>
</wsse:Security>
</soap:Header>
<soap:Body>
    <soap:Fault xmlns:wsse="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd" xmlns:c="urn:schemas-visa-com:transaction-data-1.0">
    <faultcode>wsse:FailedCheck</faultcode>
    <faultstring>
        Security Data : UsernameToken authentication failed.
    </faultstring>
</soap:Fault>
</soap:Body>
</soap:Envelope>

Answer 1

找到了一个简单的解决方案，未实现这种方式，只是一个插图：

SOAPBody body = soapResponse.getSOAPBody();
    if(body.hasFault()){
        LOG.debug("HAS LOG FAULT, LETS SEE WHAT ELSE...");
        LOG.debug("getFaultCode" + body.getFault().getFaultCode());
        LOG.debug("getFaultString" + body.getFault().getFaultString());
    }