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

when I run the marshal operation I get the following error:

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

This is my function for Marshalling...

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();

    }

Thanks for your interests..

谢谢你的兴趣..

1 个解决方案

#1

You can't marshal just a String as it doesn't have any root element information (hence the exception about the missing @XmlRootElement annotation), but you can wrap it in an instance of JAXBElement and then marshal that. JAXBElement is another way to supply this root element information to JAXB.

你不能只编组一个String，因为它没有任何根元素信息（因此关于缺少的@XmlRootElement注释的例外），但是你可以将它包装在JAXBElement的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。

Example of Creating the JAXBElement

创建JAXBElement的示例

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

If you Generated your Model From an XML Schema

如果您从XML架构生成模型

If you created your object model from an XML Schema. And you have a top-level XML element that is a data type like xs:string then there will be a convenience method on the generated ObjectFactory class that will help you create the JAXBElement instance.

如果您是从XML架构创建对象模型。并且您有一个*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您创建JAXBElement实例。

#1