Question

我有一个像这样定义的DateTimeTypeAdapter：

public class DateTimeAdapter extends XmlAdapter<String, DateTime>{
  @Override
  public DateTime unmarshal(String value) throws Exception {
    return new DateTime(value);
  }
  @Override
  public String marshal(DateTime value) throws Exception {
    return value.toString();
  }
}

包装注释： @XmlJavaTypeAdapters（{ @XmlJavaTypeAdapter（type = DateTime.class，value = DateTimeAdapter.class）}） package srm.model; import org.joda.time.DateTime;

我正试图将DateTime编组：

@Test
public void foo() throws Exception {
  Root root = new Root();
  root.dateTime = new DateTime(2011, 5, 30, 11, 2, 30, 0);
  JAXBContext jc = JAXBContext.newInstance(Root.class);

  Marshaller marshaller = jc.createMarshaller();
  marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
  marshaller.marshal(root, System.out);

  marshaller.setProperty(Marshaller.JAXB_FRAGMENT, true);
  marshaller.marshal(root, System.out);

  JAXBElement<?> element = createElement("a", DateTime.class, root.dateTime);
  marshaller.marshal(element, System.out);
}
private static <T> JAXBElement<T> createElement(String name, Class<T> type, Object value)   {
  return new JAXBElement<>(new QName(name), type, type.cast(value));
}

前两个调用成功但第三个调用抛出异常：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<root>
  <dateTime>2011-05-30T11:02:30.000-07:00</dateTime>
</root>

<root>
  <dateTime>2011-05-30T11:02:30.000-07:00</dateTime>
</root>

javax.xml.bind.MarshalException
  - with linked exception:
[com.sun.istack.internal.SAXException2: org.joda.time.DateTime is not known to this context
javax.xml.bind.JAXBException: org.joda.time.DateTime is not known to this context]
  at com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:311)
  at com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.marshal(MarshallerImpl.java:236)
  at javax.xml.bind.helpers.AbstractMarshallerImpl.marshal(AbstractMarshallerImpl.java:95)
at srm.model.XmlTest.foo(XmlTest.java:259)

如何使这项工作？谢谢！

Answer 1

XmlAdapter永远不会应用于根对象。相反，您可以转换对象，然后将转换后的对象放在JAXBElement中，然后封送它。

无法使用JAXBElement编组XmlJavaTypeAdapter？

1 个答案: