CODEWITHSUNDEEP
javaxmljaxb
Velth
Velth

Reputation: 1108

Using JAXB to create reference-objects accordingly to an attribute

Consider the following xml:

<Config>
    <Paths>
        <Path reference="WS_License"/>
    </Paths>

    <Steps>
        <Step id="WS_License" title="License Agreement" />
    </Steps>
</Config>

The following JAXB classes:

public class Path {

    private String _reference;

    public String getReference() {
        return _reference;
    }

    @XmlAttribute
    public void setReference( String reference ) {
        _reference = reference;
    }

}

And

public class Step {

    private String _id;
    private String _title;

    public String getId() {
        return _id;
    }

    @XmlAttribute
    public void setId( String id ) {
        _id = id;
    }

    public String getTitle() {
        return _title;
    }

    @XmlAttribute
    public void setTitle( String title ) {
        _title = title;
    }

}

Instead of storing the reference in the Path object as String, I'd like to hold it as a Step object. The link between those objects is the reference and id attributes. Is the @XMLJavaTypeAdapter attribute the way to go? Could anyone be so kind to provide an example of the correct usage?

Thanks!

EDIT:

I'd also would like to do the same technique with an element.

Consider the following xml:

<Config>
    <Step id="WS_License" title="License Agreement">
        <DetailPanelReference reference="DP_License" />
    </Step>

    <DetailPanels>
        <DetalPanel id="DP_License" title="License Agreement" />
    </DetailPanels>
</Config>

The following JAXB classes:

@XmlAccessorType(XmlAccessType.FIELD)
public class Step {

    @XmlID
    @XmlAttribute(name="id")
    private String _id;

    @XmlAttribute(name="title")
    private String _title;

    @XmlIDREF
    @XmlElement(name="DetailPanelReference", type=DetailPanel.class)
    private DetailPanel[] _detailPanels; //Doesn't seem to work

}

@XmlAccessorType(XmlAccessType.FIELD)
public class DetailPanel {

    @XmlID
    @XmlAttribute(name="id")
    private String _id;

    @XmlAttribute(name="title")
    private String _title;

}

The property _detailPanels in the Step-object is empty and the link doesn't seems to work. Is there any option to create a link without creating a new JAXB object holding only the reference to the DetailPanel?

Thanks again : )!

Upvotes: 3

Views: 5345

Answers (1)

bdoughan
bdoughan

Reputation: 148977

You can use @XmlID to map a property as the key and @XmlIDREF to map the reference to the key for this use case.

Step

@XmlAccessorType(XmlAccessType.FIELD)
public class Step {

    @XmlID
    @XmlAttribute
    private String _id;

}

Path

@XmlAccessorType(XmlAccessType.FIELD)
public class Path {

    @XmlIDREF
    @XmlAttribute
    private Step _reference;

}

For More Information

UPDATE

Thanks! I Completely missed your article. I've extended my question, do you have any clue if this is possible too? I do not want to create a class with only holding the reference, I'd like to store it inside the step class.

Note: I'm the EclipseLink JAXB (MOXy) lead and a member of the JAXB (JSR-222) expert group.

If you are using MOXy as your JAXB (JSR-222) provider then you could leverage the @XmlPath annotation for your use case.

import javax.xml.bind.annotation.*;
import org.eclipse.persistence.oxm.annotations.XmlPath;

@XmlAccessorType(XmlAccessType.FIELD)
public class Step {

    @XmlID
    @XmlAttribute
    private String id;

    @XmlPath("DetailPanelReference/@reference")
    @XmlIDREF
    // private List<DetailPanel> _detailPanels; // WORKS
    private DetailPanel[] _detailPanels; // See bug:  http://bugs.eclipse.org/399293

}

For More Information

Upvotes: 7

Related Questions