JAXB Unmarshalling @XmlAnyElement

Question

我创建了三个JAXB类：Home , Person , Animal。 Java类 Home有变量List<Object> any，可能包含Person和/或Animal实例。

    public class Home {
        @XmlAnyElement(lax = true)
        protected List<Object> any;
    //setter getter also implemented
    }
@XmlRootElement(name = "Person")                            // Edited
    public class Person {
        protected String name; //setter getter also implemented
     } 
@XmlRootElement(name = "Animal")                             // Edited
    public class Animal {
       protected String name; //setter getter also implemented
     }

---

/ * 解组后 * /

 Home home ;

                for(Object obj : home .getAny()){
                    if(obj instanceof Person ){
                        Person  person = (Person )obj;
                        // .........
                    }else if(obj instanceof Animal ){
                        Animal animal = (Animal )obj;
                        // .........
                    }
                }

我需要在Person or Animal变量中保存"Home.any" List个对象，但"Home.any" List的内容是com.sun.org.apache.xerces.internal.dom.ElementNSImpl的实例，而不是Animal or Person。

那么有没有办法实现Animal or Person中保存在xml中的"Home.any" List实例。

Answer 1

您需要在要使用@XmlRootElement注释的字段/属性中将要显示为实例的类添加@XmlAnyElement(lax=true)。

Java模型

主

import java.util.List;
import javax.xml.bind.annotation.*;

@XmlAccessorType(XmlAccessType.FIELD)
public class Home {
    @XmlAnyElement(lax = true)
    protected List<Object> any;

    //setter getter also implemented
}

人

import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name="Person")
public class Person {

}

<强>动物

import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name="Animal")
public class Animal {

}

演示代码

<强> input.xml中

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <Person/>
    <Animal/>
    <Person/>
</root>

<强>演示

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

public class Demo {

    public static void main(String[] args) throws JAXBException {
        JAXBContext jc = JAXBContext.newInstance(Home.class, Person.class, Animal.class);

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        StreamSource xml = new StreamSource("src/forum20329510/input.xml");
        Home home = unmarshaller.unmarshal(xml, Home.class).getValue();

        for(Object object : home.any) {
            System.out.println(object.getClass());
        }
    }

}

<强>输出

class forum20329510.Person
class forum20329510.Animal
class forum20329510.Person

更多信息

• http://blog.bdoughan.com/2012/12/jaxbs-xmlanyelementlaxtrue-explained.html