Question

for the xml:

<foo xmlns="http://ns.com"
     xmlns:ext="http://ext.com"
     attr="xxx"
     ext:bar="yyy">
</foo>

How can I create a Foo class? Specifically, I’d like to be able to separate the ‘ext’ attribute somehow so it is not directly in Foo, but in another class, and in a typesafe way (so not XmlAnyAttribute).

What I optimally wish for is:

class Foo {
  Ext ext;
}

class Ext {
  String bar;
}

Answer 1

You can map a POJO field/property with @XmlAttribute if the referenced object has only one mapped field/property with @XmlValue.

Foo

class Foo {
    @XmlAttribute(namespace="http://www.ext.com")
    Ext ext
}

Ext

class Ext {
    @XmlValue
    String bar;
}

For More Information

• http://blog.bdoughan.com/search/label/Namespaces

---

UPDATE

Note: I’m the EclipseLink JAXB (MOXy) lead and a member of the JAXB 2 (JSR-222) expert group.

What if i want to map several attributes?

You can leverage the @XmlPath extension in MOXy for this use case:

Foo

Using @XmlPath(".") indicates that you want the target object represented at the same level in the XML document as the source object.

class Foo {
    @XmlPath(".")
    Ext ext
}

Ext

class Ext {
    @XmlAttribute
    String foo;

    @XmlAttribute
    String bar;
}

For More Information

• http://blog.bdoughan.com/2010/07/xpath-based-mapping.html
• http://blog.bdoughan.com/2011/05/specifying-eclipselink-moxy-as-your.html