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JAXB,Marshal的问题, - 无法编组类型“java.lang.String”

[英]Problems with JAXB, Marshal, - unable to marshal type “java.lang.String”

当我运行marshal操作时,我收到以下错误:

javax.xml.bind.MarshalException
 - with linked exception:
[com.sun.istack.internal.SAXException2: unable to marshal type "java.lang.String" as an element because it is missing an @XmlRootElement annotation]
    ...

Caused by: com.sun.istack.internal.SAXException2: unable to marshal type "java.lang.String" as an element because it is missing an @XmlRootElement annotation
    at com.sun.xml.internal.bind.v2.runtime.XMLSerializer.reportError(XMLSerializer.java:237)
    at com.sun.xml.internal.bind.v2.runtime.LeafBeanInfoImpl.serializeRoot(LeafBeanInfoImpl.java:126)
    at com.sun.xml.internal.bind.v2.runtime.XMLSerializer.childAsRoot(XMLSerializer.java:483)
    at com.sun.xml.internal.bind.v2.runtime.MarshallerImpl.write(MarshallerImpl.java:308)
    ... 6 more

这是我编组的功能......

public StringBuffer Marshaller(Object marshall){   // make marshalling->Java to XML
        StringWriter writer = new StringWriter();
        try {
            JAXBContext jaxbContext=JAXBContext.newInstance(marshall.getClass());
            Marshaller jaxbMarshaller=jaxbContext.createMarshaller();
            // çıktı
            jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            jaxbMarshaller.marshal(marshall, writer);
            System.out.println(writer.getBuffer().toString());
        } catch (PropertyException e) {
            e.printStackTrace();
        } catch (JAXBException e) {
            e.printStackTrace();
        }
        return writer.getBuffer();

    }

谢谢你的兴趣..

你不能只编组一个String因为它没有任何根元素信息(因此有关缺少的@XmlRootElement注释的例外),但你可以将它包装在JAXBElement一个实例中,然后编组它。 JAXBElement是向JAXB提供此根元素信息的另一种方法。

创建JAXBElement示例

JAXBElement<String> jaxbElement =
  new JAXBElement(new QName("root-element"), 
    String.class, string);

如果您从XML架构生成模型

如果您是从XML Schema创建了对象模型。 你有一个顶级的XML元素,它是一个像xs:string这样的数据类型,然后在生成的ObjectFactory类上会有一个方便的方法来帮助你创建JAXBElement实例。

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