Question

当我运行编组操作时，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

这是我的编组功能......

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();

    }

感谢您的兴趣..

Answer 1

你不能只编组一个String，因为它没有任何根元素信息（因此缺少@XmlRootElement注释的例外），但你可以将它包装在JAXBElement的实例中{1}}然后封送那个。 JAXBElement是将此根元素信息提供给JAXB的另一种方法。

创建JAXBElement

的示例

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。并且您有一个顶级XML元素，它是一种数据类型，如xs:string，那么生成的ObjectFactory类将有一个方便的方法，可以帮助您创建JAXBElement实例。< / p>

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

1 个答案: