java - JAXB 解码 @XmlAnyElement

Question

我创建了三个 JAXB 类:Home , Person , Animal 。 Java类首页有变量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
     }

<小时/>

/* 解码后 */

 Home home ;

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

我需要实现Person or Animal对象保存在 "Home.any" List变量，但内容为"Home.any" List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保存在 "Home.any" List 中的 xml 中的实例.

Answer 1

您需要在想要在使用 @XmlAnyElement(lax=true) 注释的字段/属性中显示为实例的类上添加 @XmlRootElement。

Java 模型

首页

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 {

}

动物

import javax.xml.bind.annotation.XmlRootElement;

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

}

演示代码

input.xml

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

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());
        }
    }

}

输出

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

了解更多信息

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