How do I parse this XML in Java with JAXB?

Question

I have the following XML, no XSD or schema with it that I want to parse to java object(s) using JAXB as I heard its better than SAX. Is there a way to accomplish this with annotations or a better way to do this? Does it make it so that i just need a single FosterHome class? I am having trouble finding how to do this any help is grateful.

I was originally thinking of having a FosterHome, Family, and Child class. Using JAXB, is having 3 classes no longer necessary? Im not sure how to deal with this as ChildID shows up in two different areas.

<?xml version="1.0" encoding="UTF-8"?>
<FosterHome>
    <Orphanage>Happy Days Daycare</Orphanage>
    <Location>Apple Street</Location>
    <Families>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child1</ChildID>
                <ChildID>Child2</ChildID>
            </ChildList>
        </Family>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child3</ChildID>
                <ChildID>Child4</ChildID>
            </ChildList>
        </Family>
    </Families>
    <RemainingChildList>
        <ChildID>Child5</ChildID>
        <ChildID>Child6</ChildID>
    </RemainingChildList>
</FosterHome>

I'm no JAXB expert, but I believe that you need a schema of some sort for it to work. — Hovercraft Full Of Eels, Nov 18 '11 at 18:06
@Hovercraft Full of Eels: You only need a schema for validation and if you want to generate your Java bindings. — SingleShot, Nov 18 '11 at 18:15
this question is the best walk-through in the www. +1 vote for this good question. — Amir, Dec 23 '18 at 08:57

bdoughan · Accepted Answer · 2011-11-18T19:30:15.223

You could do the following. By leveraging @XmlElementWrapper you can reduce the amount of classes that you require:

FosterHome

package nov18;

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

@XmlRootElement(name="FosterHome")
@XmlAccessorType(XmlAccessType.FIELD)
public class FosterHome {

    @XmlElement(name="Orphanage")
    private String orphanage;

    @XmlElement(name="Location")
    private String location;

    @XmlElementWrapper(name="Families")
    @XmlElement(name="Family")
    private List<Family> families;

    @XmlElementWrapper(name="RemainingChildList")
    @XmlElement(name="ChildID")
    private List<String> remainingChildren;

}

Family

package nov18;

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

@XmlAccessorType(XmlAccessType.FIELD)
public class Family {

    @XmlElement(name="ParentID")
    private String parentID;

    @XmlElementWrapper(name="ChildList")
    @XmlElement(name="ChildID")
    private List<String> childList;

}

Demo

package nov18;

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

public class Demo {

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

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        FosterHome fosterHome = (FosterHome) unmarshaller.unmarshal(new File("src/nov18/input.xml"));

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

}

Input/Output

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<FosterHome>
    <Orphanage>Happy Days Daycare</Orphanage>
    <Location>Apple Street</Location>
    <Families>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child1</ChildID>
                <ChildID>Child2</ChildID>
            </ChildList>
        </Family>
        <Family>
            <ParentID>Adams</ParentID>
            <ChildList>
                <ChildID>Child3</ChildID>
                <ChildID>Child4</ChildID>
            </ChildList>
        </Family>
    </Families>
    <RemainingChildList>
        <ChildID>Child5</ChildID>
        <ChildID>Child6</ChildID>
    </RemainingChildList>
</FosterHome>

For More Information

http://blog.bdoughan.com/2010/09/jaxb-collection-properties.html

UPDATE

Is there I easy way I can iterate/print out all the ChildID in the Family class?

You could do the following:

    for(Family family : fosterHome.getFamilies()) {
        System.out.println(family.getParentID());
        for(String childID : family.getChildList()) {
            System.out.println("    " + childID);
        }
    }

Is there I easy way I can iterate/print with System.out all the ChildID in the Family class within the main method? — Rolando, Nov 18 '11 at 19:25
@Tiever - I have updated my answer with how this can be done. — bdoughan, Nov 18 '11 at 19:30
@BlaiseDoughan excellent solution. Even back in the 2011 days...However how to get ChildID of the type FosterHome, it looks like ChildID is of the type Family, so if you wanted the fosterHome object to contain the childIDs, this may not work. Do you know of a way to accomplish this? Something like `FosterHome.Family` may not be possible since the classes are decoupled. — mosawi, Apr 04 '17 at 01:10

Abhishek · Answer 2 · 2011-11-18T19:27:46.540

try {
    // create a JAXBContext capable of handling classes generated into
    // the com.abhi.xml.jaxb.generated package
    JAXBContext jc = JAXBContext.newInstance( "com.abhi.xml.jaxb.generated" );

    // create an Unmarshaller
    Unmarshaller u = jc.createUnmarshaller();

    // unmarshal a FosterHome instance document into a tree of Java content
    // objects composed of classes from the com.abhi.xml.jaxb.generated 
    // package.
    JAXBElement<?> fhElement =(JAXBElement<?>)u.unmarshal
    (new FileInputStream("yourfile.xml"));
    FosterHome FH = (FosterHome)fhElement.getValue();
    System.out.println(FH.getDesc());
         // so on ..you can get all elements based on generated objects

} catch( JAXBException je ) {
    je.printStackTrace();
} catch( IOException ioe ) {
    ioe.printStackTrace();
}

Here is an example of how it can be done when staring from classes: http://stackoverflow.com/questions/8186896/how-do-i-parse-this-xml-in-java-with-jaxb/8187505#8187505 — bdoughan, Nov 18 '11 at 19:07

score 1 · Answer 3 · answered Nov 18 '11 at 18:15

1

This seems to be a good tutorial for doing this

http://blog.espenberntsen.net/2010/02/26/generate-jaxb-classes-from-an-xsd-schema-and-the-schema-from-an-xml-sample-document/

It details how to generate an xsd schema from and xml file and then how to use that schema to genenate jaxb classes. In the end you should end up with more than one class.

answered Nov 18 '11 at 18:15

Danny

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Here is an example of how it can be done when staring from classes: http://stackoverflow.com/questions/8186896/how-do-i-parse-this-xml-in-java-with-jaxb/8187505#8187505 – bdoughan Nov 18 '11 at 19:06

How do I parse this XML in Java with JAXB?

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