使用 Java 从无效的 SOAP 1.1 消息中获取值

Question

我的 previous question 已关闭并标记为重复，但建议的 asnwer 没有回答我的问题，而且按照建议，我在问一个新问题。

让我们使用建议的。

代码如下：

String strMsg = "<soap:Envelope xmlns:soap=\"http://www.w3.org/2003/05/soap-envelope\">" +
              " <soap:Header>" +
              "  <context xmlns=\"urn:zimbra\"/>" +
              " </soap:Header>" +
              " <soap:Body>" +
              "  <soap:Fault>" +
              "   <soap:Code>" +
              "    <soap:Value>soap:Sender</soap:Value>" +
              "   </soap:Code>" +
              "   <soap:Reason>" +
              "    <soap:Text>no valid authtoken present</soap:Text>" +
              "   </soap:Reason>" +
              "   <soap:Detail>" +
              "    <Error xmlns=\"urn:zimbra\">" +
              "     <Code>service.AUTH_REQUIRED</Code>" +
              "     <Trace>qtp1027591600-6073:1588614639199:4eacbd0257a457b6</Trace>" +
              "    </Error>" +
              "   </soap:Detail>" +
              "  </soap:Fault>" +
              " </soap:Body>" +
              "</soap:Envelope>";

      InputStream is = new ByteArrayInputStream(strMsg.getBytes());
      XMLInputFactory xif = XMLInputFactory.newFactory();
      XMLStreamReader xsr = xif.createXMLStreamReader(is);

      //Envelope
      xsr.nextTag();
      QName name = xsr.getName();

      //Header
      xsr.nextTag();
      name = xsr.getName();

      //Context
      xsr.nextTag();
      name = xsr.getName();

      //Context again
      xsr.nextTag();
      name = xsr.getName();

      //Header again
      xsr.nextTag();
      name = xsr.getName();

      //Body
      xsr.nextTag();
      name = xsr.getName();

      //Fault
      xsr.nextTag();
      name = xsr.getName();

      /* IM COMMENTING THE FOLLOWING CODE BECAUSE I'M INTERESTED IN THE FAULT CONTENT
      * AND EVEN IF IT TRY TO GO DEEPER I CAN'T GO PASS "VALUE" NODE
      *
      //Code
      xsr.nextTag();
      name = xsr.getName();

      //Value
      xsr.nextTag();
      name = xsr.getName();

      //throws exception, no more elements for some reason
      xsr.nextTag();
      name = xsr.getName();
       */

      Transformer transformer = TransformerFactory.newInstance().newTransformer();
      StringWriter stringWriter = new StringWriter();
      transformer.transform(new StAXSource(xsr), new StreamResult(stringWriter));
      StringReader sr = new StringReader(stringWriter.toString());
      JAXBContext jaxbContext = JAXBContext.newInstance(Fault.class);
      Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();

      Fault fault = (Fault) unmarshaller.unmarshal(sr); //THROWS EXCEPTION
      //"unexcpected element (URI:"http://www.w3.org/2003/05/soap-envelope", local:"Fault"). Expected elements are <{}Fault>

我的错class：

  @XmlRootElement(name = "Fault")
  @XmlAccessorType(XmlAccessType.FIELD)
  public static class Fault {
    @XmlElement(name = "Code")
    private String code;
    @XmlElement(name = "Reason")
    private String reason;

    public String getCode() {
      return code;
    }

    public void setCode(String code) {
      this.code = code;
    }

    public String getReason() {
      return reason;
    }

    public void setReason(String reason) {
      this.reason = reason;
    }
  }

我怀疑它不会起作用，因为直接在 "Fault" 中的元素本身没有值，它们内部有更多元素，而且所有元素都以 "soap" 为前缀，我的信封与建议答案中的信封结构不完全相同。

但是，我仍然无法获取 "Fault" 节点，因为抛出了异常：

意外元素（URI：“http://www.w3.org/2003/05/soap-envelope”，本地："Fault"）。预期元素为 <{}Fault>

我有兴趣获取以下值：

<soap:Text>no valid authtoken present</soap:Text>"

此外，这仅适用于此类错误，可能还有其他错误，而且，当答案是肯定的时，我得到完全不同的响应。

我真正感兴趣的是，通过以下方式找到一种探索信封的方法：

//pseudo code
(envelope->body->fault->reason->text != null) {reasonText = envelope->body->fault->reason->text)

但是无论我以何种方式到达 Reason->Text 都可以，然后我可以将脚本改编为其他主体。

提前致谢。

Answer 1

我的一个朋友不想在这里回答，找到了解决办法，更好的是，他用我想要的方式找到了：

//pseudo code
(envelope->body->fault->reason->text != null) {reasonText = envelope->body->fault->reason->text)

对于未来偶然发现这个问题的任何其他人，这是一个解决方案：

String strMsg = "<soap:Envelope xmlns:soap=\"http://www.w3.org/2003/05/soap-envelope\">" +
              " <soap:Header>" +
              "  <context xmlns=\"urn:zimbra\"/>" +
              " </soap:Header>" +
              " <soap:Body>" +
              "  <soap:Fault>" +
              "   <soap:Code>" +
              "    <soap:Value>soap:Sender</soap:Value>" +
              "   </soap:Code>" +
              "   <soap:Reason>" +
              "    <soap:Text>no valid authtoken present</soap:Text>" +
              "   </soap:Reason>" +
              "   <soap:Detail>" +
              "    <Error xmlns=\"urn:zimbra\">" +
              "     <Code>service.AUTH_REQUIRED</Code>" +
              "     <Trace>qtp1027591600-6073:1588614639199:4eacbd0257a457b6</Trace>" +
              "    </Error>" +
              "   </soap:Detail>" +
              "  </soap:Fault>" +
              " </soap:Body>" +
              "</soap:Envelope>";

      strMsg = strMsg.replaceAll("soap:",""); //Had to replace soap:, not fancy but it works.

      is = new ByteArrayInputStream(strMsg.getBytes());
      InputSource xml = new InputSource(is);

      XPath xPath = XPathFactory.newInstance().newXPath();
      Object exprEval = xPath.compile("/Envelope/Body/Fault/Reason/Text/text()").evaluate(xml, XPathConstants.STRING);
      if ( exprEval != null ) {
        System.out.println( "Fault reason text: " + exprEval );
        // This prints what's expected:
        // Fault reason text: no valid authtoken present
      }

给你。

使用 Java 从无效的 SOAP 1.1 消息中获取值

Get value from non valid SOAP 1.1 Message with Java

java

soap

web-services