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如何使用JAXB保留XML注释？

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如何在不丢失注释的情况下解组和编组XML文件？有没有办法使用JAXB，我尝试使用以下link但它不起作用 .

<customer>
        <address>
    <!--   comments line 1 -->
           <street>1 Billing Street</street>
        </address>
        <address>
    <!-- comments line 2-->
        <street>2 Shipping Road</street>
       </address>
</customer>

我想解组上面的xml，为它添加一个新地址并对其进行编组，而不会丢失以下注释 .

<!--   comments line 1 -->
   <!-- comments line 2-->
java jaxb marshalling unmarshalling

2 回答

  • 0

    也许更容易调整/扩展客户地址的（数据）模型 .

    <customer>
  <billing-address>
    <street></street>
    <street></street>
    <city></city>
  </billing-address>
  <shipping-address>
    <street></street>
    <street></street>
    <street></street>
    <city></city>
  </shipping-address>
</customer>

    这样，语义在模型中不会在注释中“隐藏” .

    回复于
  • 0

    您可以将JAXB与StAX结合使用以访问尾随注释 .

    import javax.xml.bind.JAXBContext;
import javax.xml.bind.Unmarshaller;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamConstants;
import javax.xml.stream.XMLStreamReader;
import javax.xml.transform.stream.StreamSource;

public class Demo {

    public static void main(String[] args) throws Exception {
        XMLInputFactory xif = XMLInputFactory.newFactory();
        StreamSource source = new StreamSource("pathOfYourXML/input.xml");
        XMLStreamReader xsr = xif.createXMLStreamReader(source);

        JAXBContext jc = JAXBContext.newInstance(Customer.class);
        Unmarshaller unmarshaller = jc.createUnmarshaller();
        Customer xml = (Customer) unmarshaller.unmarshal(xsr);

        while(xsr.hasNext()) {
            if(xsr.getEventType() == XMLStreamConstants.COMMENT) {
                System.out.println(xsr.getText());
            }
            xsr.next();
        }
    }

}
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