Question

让我们说，我有以下课程：

HTML("""<a role="button" class="btn btn-default"
    href="{% url "some_cancel_url" %}">Cancel</a>"""),

如果some_cancel_url - 元素是＆＃34;请求＆＃34;否则 - 元素是＆＃34;响应＆＃34;。当一个请求时，该元素应被（un）封送到（来自）名为@XmlRootElement class Payment { @XmlElement int amount; @XmlElement Currency currency; @XmlElement IResponse response; }的根元素，当一个响应 - 来自（来自）response==null。

如何配置这样的编组算法？如果JAXB无法做到，可能还有其他一些引擎可以吗？

Answer 1

像这样创建和编组根元素：

JAXBElement<Payment> jbe;
if( payment.getResponse() != null ){
    jbe = wrap( null, "PaymentResponse", payment );
} else {
    jbe = wrap( null, "PaymentRequest", payment );
}
m.marshal( jbe, sw );

使用简单的辅助方法

<T> JAXBElement<T> wrap( String ns, String tag, T o ){
    QName qtag = new QName( ns, tag );
    Class<?> clazz = o.getClass();
    @SuppressWarnings( "unchecked" )
    JAXBElement<T> jbe = new JAXBElement( qtag, clazz, o );
    return jbe;
}

使解组成为可能的一种简单方法是创建两个子类PaymentResponse和PaymentRequest，它们充当@XmlRootElements。 ObjectFactory包含

@XmlElementDecl(namespace = "", name = "PaymentRequest")
public JAXBElement<PaymentRequest> 
createPaymentRequest(PaymentRequest value) {
    return new JAXBElement<PaymentRequest>(_Payment_QNAME, PaymentRequest.class, null, value);
}
@XmlElementDecl(namespace = "", name = "PaymentResponse")
public JAXBElement<PaymentResponse> 
createPaymentResponse(PaymentResponse value) {
    return new JAXBElement<PaymentResponse>(_Payment_QNAME, PaymentResponse.class, null, value);
}

解组：

JAXBContext jc = JAXBContext.newInstance( PACKAGE );
Unmarshaller m = jc.createUnmarshaller();
JAXBElement<?> tb = null;
try {
    Payment payment = readFrom( Payment.class );
} catch( Exception e  ){
}

和方法readFrom：

public <T> T readFrom( Class<T> type ) throws Exception {
    try {
        JAXBContext jc = JAXBContext.newInstance( PACKAGE );
        Unmarshaller u = jc.createUnmarshaller();
        JAXBElement<T> jbe = (JAXBElement<T>)u.unmarshal( new File( XMLIN ) );
        return type.cast( jbe.getValue() );
    } catch (JAXBException e) {
        e.printStackTrace();
    }
    return null;
}

Answer 2

我终于通过拦截StAX事件来实施解编。这是代码：

JAXBContext jc = JAXBContext.newInstance(RootElement.class, A.class, B.class, C.class, D.class, E.class);
Unmarshaller unmarsh = jc.createUnmarshaller();
XMLStreamReader xs = XMLInputFactory.newInstance().createXMLStreamReader(new StringReader(etalonRs));
XMLStreamReader xd = new StreamReaderDelegate(xs) {
      public static final String ROOT_ELEMENT = "TestRoot";
      public static final int REPLACEABLE_LEVEL = 2;
      public final Collection<String> sufficesToDelete = Arrays.asList("Rq", "Rs");

      protected Stack<String> elementNamesStack = new Stack<>();
      protected Set<String> replaceableNames = new HashSet<>();

      @Override
      public String getLocalName() {
          String realName = super.getLocalName();
          if (replaceableNames.contains(realName) && elementNamesStack.size() == REPLACEABLE_LEVEL) {
              for (String suffix : sufficesToDelete) {
                  if (realName.endsWith(suffix)) {
                      return realName.substring(0, realName.lastIndexOf(suffix));
                  }
              }
          }
          return realName;
      }

      @Override
      public int next() throws XMLStreamException {
          final int eventCode = super.next();
         processLevel(eventCode);
          return eventCode;
      }

      @Override
      public int nextTag() throws XMLStreamException {
          final int eventCode = super.nextTag();
          processLevel(eventCode);
          return eventCode;
      }

      private void processLevel(int eventCode) {
          switch (eventCode) {
              case XMLStreamReader.START_ELEMENT:
                  final String origElementName = super.getLocalName();
                  if ((elementNamesStack.size() + 1) == REPLACEABLE_LEVEL && elementNamesStack.peek().equals(ROOT_ELEMENT))
                      replaceableNames.add(origElementName);
                  elementNamesStack.push(origElementName);
                  break;
              case XMLStreamReader.END_ELEMENT: 
                  assert(elementNamesStack.pop().equals(super.getLocalName()));
                  break;

          }
      }
  };

Object o = unmarsh.unmarshal(xd);

这是我的测试课程。是的，生产中的真实结构更复杂 - 有不同的付款和＃34;并且它们的元素不在根中，所以我必须使用@XmlAnyElement注释：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "TestRoot")
public static class RootElement {
    @XmlElement(name = "SomeDate")
    private Date dt = new Date();

    @XmlAnyElement(lax=true)
    private A a = new C();
}

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "TestA")
@XmlType
public static abstract class A {
    private int fld1 = 1;

    @XmlAnyElement(lax=true)
    @XmlElementWrapper
    protected List<Object> list = new ArrayList<>();
}

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement(name = "TestC")
public static class C extends A {
    private int fld2  = 3;
}

Marshaling可以以相同的方式实现，但您必须编写＆＃34; StreamWriterDelegate＆＃34;从头开始。

JAXB：将一个类中的不同XML元素编组/解组

2 个答案: