Question

我有以下类型，在架构中定义为：

<xsd:complexType name="any_t" mixed="true">
  <xsd:sequence>
    <xsd:any processContents="lax" minOccurs="0" maxOccurs="unbounded"></xsd:any>
  </xsd:sequence>
</xsd:complexType>

生成的JAXB类：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "any_t", propOrder = {
    "content"
})
public class AnyT implements Serializable
{
    private final static long serialVersionUID = 1L;
    @XmlMixed
    @XmlAnyElement(lax = true)
    protected List<Object> content;
    public List<Object> getContent() {
        if (content == null) {
            content = new ArrayList<Object>();
        }
        return this.content;
    }
}

使用此类型的一些JAXB类：

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "", propOrder = {
    "addInfo"
})
@XmlRootElement(name = "FORM_OF_COOWNERS")
public class FormOfCoowners
    implements Serializable
{
    private final static long serialVersionUID = 1L;
    @XmlElement(name = "add_info")
    protected AnyT addInfo;
    public AnyT getAddInfo() { return addInfo; }
    public void setAddInfo(AnyT value) { this.addInfo = value; }
}

当我将此XML解组为JAXB对象时：

<FORM_OF_COOWNERS>
   <add_info>
      <info_analytics>
         <issuer_subdiv>
            <id><id>1940001</id></id>
         </issuer_subdiv>
         <fl_worker>Yes</fl_worker>
      </info_analytics>
   </add_info>
</FORM_OF_COOWNERS>

并将此对象编组为JSON，我得到：

{"FORM_OF_COOWNERS":{"add_info":{"info_analytics":[{"issuer_subdiv":{"id":{"id":"1940001"}},"fl_worker":"Yes"}]}}}

此列表来自哪里？ info_analytics映射到没有任何集合/数组的类型。

如果我得到这个JSON，将其转换为JAXB对象并编组为XML，它会生成：

<info_analytics fl_worker="Yes">
  <issuer_subdiv>
    <id id="1940001">
       <id>1940001</id>
    </id>
  </issuer_subdiv>
  <fl_worker>Yes</fl_worker>
</info_analytics>

为什么元素作为属性重复？

更新：编组/解组json：

def unmarshallToJaxbTyped[A : ClassTag](json: String): Throwable \/ A =
  \/.fromTryCatchNonFatal {
    val cls = classTag[A].runtimeClass
    val jc = JAXBContext.newInstance(cls)
    val unmarshaller = jc.createUnmarshaller()
    unmarshaller.setProperty("eclipselink.media-type", "application/json")
    unmarshaller.unmarshal(new StreamSource(new StringReader(json)), cls)
      .getValue.asInstanceOf[A]
  }

def marshallJaxbToJson(obj: Any): Throwable \/ String =
  \/.fromTryCatchNonFatal {
    val jc = JAXBContext.newInstance(obj.getClass)
    val marshaller = jc.createMarshaller()
    marshaller.setProperty("eclipselink.media-type", "application/json")
    val writer = new StringWriter()
    marshaller.marshal(obj, writer)
    writer.getBuffer.toString
  }

的xml：

protected static synchronized JAXBContext getContext() {
  if (jaxbContext == null)
    try {
      jaxbContext = JAXBContext.newInstance("org.example.models");
    } catch (JAXBException ex) {
      throw new RuntimeException(ex);
    }
  return jaxbContext;
}

public static void saveMessage(Object msg, OutputStream os) throws JAXBException {
  Marshaller marshaller = getContext().createMarshaller();
  marshaller.marshal(msg, os);
}

public static Object loadMessage(InputStream is) throws JAXBException {
  Unmarshaller unmarshaller = getContext().createUnmarshaller();
  return unmarshaller.unmarshal(is);
}

Answer 1

我不确定如此奇怪的往返：XML - ＆gt; Java - ＆gt; JSON - ＆gt; Java - ＆gt;需要XML。我想你明白，XML支持比JSON更多的功能，这就是为什么并不总是能够进行1对1的映射。

如果这对您来说是一个真正的问题而且不仅仅是理论研究，请针对MOXy发出错误：https://bugs.eclipse.org/bugs/enter_bug.cgi?product=EclipseLink

奇怪的XmlMixed XML＆lt; - ＆gt; JAXB＆lt; - ＆gt;使用MOXy进行JSON编组/解组

1 个答案: