Unmarshalling an XML using Xpath expression and jaxb

问题

I am new to JAXB and would like to know if there is a way by which I can unmarshall an XML to my response object but using xpath expressions. The issue is that I am calling a third party webservice and the response which I receive has a lot of details. I do not wish to map all the details in the XML to my response object. I just wish to map few details from the xml using which I can get using specific XPath expressions and map those to my response object. Is there an annotation which can help me achieve this?

For example consider the following response

<root>
  <record>
    <id>1</id>
    <name>Ian</name>
    <AddressDetails>
      <street> M G Road </street>
    </AddressDetails>
  </record>  
</root>

I am only intrested in retrieving the street name so I want to use xpath expression to get value of street using 'root/record/AddressDetails/street' and map it to my response object

public class Response{
     // How do i map this in jaxb, I do not wish to map record,id or name elements
     String street; 

     //getter and setters
     ....
}   

Thanks

回答1:

Note: I'm the EclipseLink JAXB (MOXy) lead and a member of the JAXB (JST-222) expert group.

You could use MOXy's @XmlPath extension for this use case.

Response

import javax.xml.bind.annotation.*;
import org.eclipse.persistence.oxm.annotations.XmlPath;

@XmlRootElement(name="root")
@XmlAccessorType(XmlAccessType.FIELD)
public class Response{
    @XmlPath("record/AddressDetails/street/text()")
    String street; 

    //getter and setters
}

jaxb.properties

To use MOXy as your JAXB provider you need to include a file called jaxb.properties in the same package as your domain model with the following entry (see: http://blog.bdoughan.com/2011/05/specifying-eclipselink-moxy-as-your.html)

javax.xml.bind.context.factory=org.eclipse.persistence.jaxb.JAXBContextFactory

Demo

import java.io.File;
import javax.xml.bind.*;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(Response.class);

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        File xml = new File("src/forum17141154/input.xml");
        Response response = (Response) unmarshaller.unmarshal(xml);

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(response, System.out);
    }

}

Output

<?xml version="1.0" encoding="UTF-8"?>
<root>
   <record>
      <AddressDetails>
         <street> M G Road </street>
      </AddressDetails>
   </record>
</root>

For More Information

• http://blog.bdoughan.com/2010/07/xpath-based-mapping.html
• http://blog.bdoughan.com/2011/05/specifying-eclipselink-moxy-as-your.html

回答2:

If all you want is the street name, just use an XPath expression to get it as a string, and forget about JAXB - the complex JAXB machinery is not adding any value.

import javax.xml.xpath.*;
import org.xml.sax.InputSource;

public class XPathDemo {

    public static void main(String[] args) throws Exception {
        XPathFactory xpf = XPathFactory.newInstance();
        XPath xpath = xpf.newXPath();

        InputSource xml = new InputSource("src/forum17141154/input.xml");
        String result = (String) xpath.evaluate("/root/record/AddressDetails/street", xml, XPathConstants.STRING);
        System.out.println(result);
    }

}

回答3:

If you want to map to the middle of an XML document you could do the following:

Demo Code

You could use a StAX parser to advance to the part of the document you wish to unmarshal, and then unmarshal the XMLStreamReader from there,.

import javax.xml.bind.*;
import javax.xml.stream.*;
import javax.xml.transform.stream.StreamSource;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(Response.class);

        XMLInputFactory xif = XMLInputFactory.newFactory();
        StreamSource xml = new StreamSource("src/forum17141154/input.xml");
        XMLStreamReader xsr = xif.createXMLStreamReader(xml);
        while(!(xsr.isStartElement() && xsr.getLocalName().equals("AddressDetails"))) {
            xsr.next();
        }

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        JAXBElement<Response> response = unmarshaller.unmarshal(xsr, Response.class);

        Marshaller marshaller = jc.createMarshaller();
        marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
        marshaller.marshal(response, System.out);
    }

}

Response

The mappings on the domain object are made relative to the fragment of the XML document you are mapping to.

import javax.xml.bind.annotation.*;

@XmlAccessorType(XmlAccessType.FIELD)
public class Response{

    String street; 

    //getter and setters
}

Output

<?xml version="1.0" encoding="UTF-8"?>
<AddressDetails>
   <street> M G Road </street>
</AddressDetails>

For More Information

• http://blog.bdoughan.com/2012/08/handle-middle-of-xml-document-with-jaxb.html

来源：https://stackoverflow.com/questions/17141154/unmarshalling-an-xml-using-xpath-expression-and-jaxb