JAXB Unmarshalling @XmlAnyElement
Asked Answered
A

1

6

I have created three JAXB class : Home , Person , Animal . Java Class Home have variable List<Object> any that may contain Person and/or Animal instance .

    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
     }

/* After Unmarshalling */

 Home home ;

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

I need to achieve Person or Animal object saved in "Home.any" List variable but content of "Home.any" List is instance of com.sun.org.apache.xerces.internal.dom.ElementNSImpl instead of Animal or Person .

So is there a way to achieve Animal or Person instance that is saved in xml in "Home.any" List.

Apery answered 2/12, 2013 at 13:29 Comment(0)
K
8

You need to add @XmlRootElement on the classes you want to appear as instances in the field/property you have annotated with @XmlAnyElement(lax=true).

Java Model

Home

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
}

Person

import javax.xml.bind.annotation.XmlRootElement;

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

}

Animal

import javax.xml.bind.annotation.XmlRootElement;

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

}

Demo Code

input.xml

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

Demo

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());
        }
    }

}

Output

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

For More Information

Kaisership answered 2/12, 2013 at 13:48 Comment(8)
I have already added @XmlRootElement(name = "Person") and @XmlRootElement(name = "Animal") also added entry in ObjectFactory.java .Apery
@userG - How are you creating your JAXBContext and is it aware of all your classes?Kaisership
According to your post , i think ; there is a problem with JAXBContext Declaration JAXBContext.newInstance(Home.class) .Apery
I have already defined both @XmlRootElement and entry in ObjectFactory.java . Is i have to define only one of them ?Apery
JAXBContext.newInstance(Home.class,Person.class,Animal.class) , @XmlRootElement already added and entry from ObjectFactory.java commented but still not resolved .Apery
@userG - I have updated my answer with a complete code example.Kaisership
Thanks @Blaise for your guidance , tested your code working fine as stated . I am working on netbeans plugin and using api org.netbeans.modules.xml.jaxb.api , i am still facing this problem and for temporary solution , passing dom.Element object to unmarshall method again to retrieve jaxb object . (I have also seen JAXBContext.typeMap Person & Animal both reference exist .)Apery
What about java.lang.* classes like String? Any tips on how to unmarshall a List<String>?Faradism

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