如何使用JAXB从服务返回的“ anyType”中创建Java对象?


问题内容

Web服务正在返回由WSDL定义的对象,该对象为:

<s:complexType mixed="true"><s:sequence><s:any/></s:sequence></s:complexType>

当我打印出该对象的类信息时,它显示为:

class com.sun.org.apache.xerces.internal.dom.ElementNSImpl

但我需要将此对象解组为以下类的对象:

@XmlAccessorType(XmlAccessType.FIELD) 
@XmlType(name = "", propOrder = {
        "info",
        "availability",
        "rateDetails",
        "reservation",
        "cancellation",
        "error" }) 
@XmlRootElement(name = "ArnResponse") 
public class ArnResponse { }

我知道响应是正确的,因为我知道如何编组此对象的XML:

Marshaller m = jc.createMarshaller();
m.setProperty( Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE );
m.marshal(rootResponse, System.out);

打印出:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ns2:SubmitRequestDocResponse xmlns:ns2="http://tripauthority.com/hotel">
    <ns2:SubmitRequestDocResult>
        <!-- below is the object I'm trying to unmarshall -->
        <ArnResponse>
            <Info />
            <Availability>
                <!-- etc--> 
             </Availability>
        </ArnResponse>
    </ns2:SubmitRequestDocResult>
</ns2:SubmitRequestDocResponse>

如何将ElementNSImpl看到的ArnResponse对象变成我知道的对象?

此外,我在AppEngine上运行,该文件访问受到限制。

谢谢你的帮助

更新

我添加了@XmlAnyElement(lax=true)注释,如下所示:

  @XmlAccessorType(XmlAccessType.FIELD)
  @XmlType(name = "", propOrder = {
      "content"
  })
  @XmlSeeAlso(ArnResponse.class)
  public static class SubmitRequestDocResult {

    @XmlMixed
    @XmlAnyElement(lax = true)
    protected List<Object> content;

但这没有任何区别。

这与内容是的事实有关List吗?

这是从服务器取回内容后要尝试访问的代码:

List list = rootResponse.getSubmitRequestDocResult().getContent();

for (Object o : list) {
  ArnResponse response = (ArnResponse) o;
  System.out.println(response);
}

具有以下输出:

2012年1月31日,上午10:04:14 com.districthp.core.server.ws.alliance.AllianceApi
getRates严重:com.sun.org.apache.xerces.internal.dom.ElementNSImpl无法转换为com.districthp.core
.server.ws.alliance.response.ArnResponse

回答:

axtavt的答案就解决了。这工作:

Object content = ((List)result.getContent()).get(0);
JAXBContext context = JAXBContext.newInstance(ArnResponse.class);
Unmarshaller um = context.createUnmarshaller();
ArnResponse response = (ArnResponse)um.unmarshal((Node)content);
System.out.println("response: " + response);

问题答案:

您可以将该对象传递给Unmarshaller.unmarshal(Node),它应该可以将其解组。