使用Xpath表达式和jaxb解组XML

Question

我是JAXB的新手,想知道是否有一种方法可以将XML解组为我的响应对象但使用xpath表达式.问题是我打电话给第三方网络服务,我收到的回复有很多细节.我不希望将XML中的所有细节映射到我的响应对象.我只希望从xml中映射一些细节,使用特定的XPath表达式将其映射到我的响应对象.是否有注释可以帮助我实现这一目标？

例如,考虑以下响应

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

我只对检索街道名称感兴趣所以我想使用xpath表达式使用'root/record/AddressDetails/street'获取street的值并将其映射到我的响应对象

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
     ....
}   

谢谢

Answer 1

注意: 我是EclipseLink JAXB(MOXy)的负责人,也是JAXB(JST-222)专家组的成员.

您可以@XmlPath在此用例中使用MOXy的扩展名.

响应

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

要将MOXy用作JAXB提供程序,您需要jaxb.properties在与域模型相同的程序包中包含一个名为的文件,并带有以下条目(请参阅:http: //blog.bdoughan.com/2011/05/specifying-eclipselink-moxy-as -your.html)

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

演示

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);
    }

}

产量

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

欲获得更多信息

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

Answer 2

如果你想要的只是街道名称,只需使用XPath表达式将其作为字符串,并忘记JAXB - 复杂的JAXB机制没有添加任何值.

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);
    }

}