以编程方式添加或设置JAXB XmlRootElement

Question

我想避免向每个POJO添加注释@XMLRootElement，我将使用它来解组xml。是否有一种方法来设置或以编程方式添加XmlRootElement类?

 //@XmlRootElement here
    public class Person{
    private String name;
    private String address;
    //other fields
    //getters and setters here

如果不可能，我怎么能解编组一个xml没有XmlRootElement在POJO?

public static <T> T unmarshal(Class clazz, String xml) {
    try {
        JAXBContext ctx = JAXBContext.newInstance(clazz);
        Unmarshaller u = ctx.createUnmarshaller();
        return (T) u.unmarshal(new StringReader(xml));
    } catch (JAXBException e) {
        throw new RuntimeException("Error interpreting XML response", e);
    }
}

没有@XmlRootElement就抛出异常:

Caused by: javax.xml.bind.UnmarshalException: unexpected element (uri:"", local:"report"). Expected elements are (none)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext.handleEvent(UnmarshallingContext.java:556)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportError(Loader.java:199)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportError(Loader.java:194)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.Loader.reportUnexpectedChildElement(Loader.java:71)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext$DefaultRootLoader.childElement(UnmarshallingContext.java:962)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext._startElement(UnmarshallingContext.java:399)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallingContext.startElement(UnmarshallingContext.java:380)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.SAXConnector.startElement(SAXConnector.java:101)
at org.apache.xerces.parsers.AbstractSAXParser.startElement(Unknown Source)
at org.apache.xerces.impl.XMLNSDocumentScannerImpl.scanStartElement(Unknown Source)
at org.apache.xerces.impl.XMLNSDocumentScannerImpl$NSContentDispatcher.scanRootElementHook(Unknown Source)
at org.apache.xerces.impl.XMLDocumentFragmentScannerImpl$FragmentContentDispatcher.dispatch(Unknown Source)
at org.apache.xerces.impl.XMLDocumentFragmentScannerImpl.scanDocument(Unknown Source)
at org.apache.xerces.parsers.XML11Configuration.parse(Unknown Source)
at org.apache.xerces.parsers.XML11Configuration.parse(Unknown Source)
at org.apache.xerces.parsers.XMLParser.parse(Unknown Source)
at org.apache.xerces.parsers.AbstractSAXParser.parse(Unknown Source)
at org.apache.xerces.jaxp.SAXParserImpl$JAXPSAXParser.parse(Unknown Source)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal0(UnmarshallerImpl.java:195)
at com.sun.xml.internal.bind.v2.runtime.unmarshaller.UnmarshallerImpl.unmarshal(UnmarshallerImpl.java:168)
at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:137)
at javax.xml.bind.helpers.AbstractUnmarshallerImpl.unmarshal(AbstractUnmarshallerImpl.java:194)

编辑:

布莱斯的回答确实起作用了。

StreamSource source = new StreamSource(new StringReqder(xml));
return (T) u.unmarshal(source, clazz).getValue();

但是当我试图用它来返回List

时

//this works when I specify @XmlRootElement
private static <T> List<T> unmarshallCollection(Class<T> clazz, Source source)
            throws JAXBException {
        JAXBContext ctx = JAXBContext.newInstance(WrapperCollection.class, clazz);
        Unmarshaller u = ctx.createUnmarshaller();
        WrapperCollection<T> collection = u.unmarshal(source, WrapperCollection.class).getValue();
        return collection.getItems();
    }

会抛出一个异常org.apache.xerce .dom. elementnsimpl不能被强制转换为。我做错了什么?

Answer 1

您只需要使用接受Class参数的unmarshal方法之一。

StreamSource source = new StreamSource(new StringReqder(xml));
return (T) u.unmarshal(source, clazz).getValue();

注意

• 解组方法的结果将是JAXBElement的一个实例，它包含根元素信息，您可以通过调用getValue来获得解组对象。
• 对于封送，你可以通过在JAXBElement的实例中包装你的对象来提供根信息，然后封送。