How to generate a Java class which implements Serializable interface from xsd using JAXB?

Boris Pavlović picture Boris Pavlović · Oct 3, 2009 · Viewed 53.8k times · Source

I would like to introduce caching into an existing Spring project which uses JAXB to expose WebServices. Caching will be done on the level of end points. In order to do that classes generated from XSD using JAXB need to implement Serializable interface and override Object's toString() method.

How to instruct the xjc tool using XSD to generate source with needed properties?

Answer

Pascal Thivent picture Pascal Thivent · Oct 3, 2009

Serializable

Use xjc:serializable in a custom bindings file to add the java.io.Serializable interface to your classes along with a serialVersionUID:

<?xml version="1.0" encoding="UTF-8"?>
<bindings xmlns="http://java.sun.com/xml/ns/jaxb"
            xmlns:xsi="http://www.w3.org/2000/10/XMLSchema-instance"
            xmlns:xjc="http://java.sun.com/xml/ns/jaxb/xjc"
            xsi:schemaLocation="
http://java.sun.com/xml/ns/jaxb http://java.sun.com/xml/ns/jaxb/bindingschema_2_0.xsd"
version="2.1">
  <globalBindings>
    <serializable uid="1" />
  </globalBindings>
</bindings> 

toString()

Use a superclass (see xjc:superClass) from which all your bound classes will inherit. This class won’t be generated by xjc so you are free to create it as you please (here with a toString() implementation):

<?xml version="1.0" encoding="UTF-8"?>
<bindings xmlns="http://java.sun.com/xml/ns/jaxb"
                xmlns:xsi="http://www.w3.org/2000/10/XMLSchema-instance"
                xmlns:xjc="http://java.sun.com/xml/ns/jaxb/xjc"
                xsi:schemaLocation="
http://java.sun.com/xml/ns/jaxb http://java.sun.com/xml/ns/jaxb/bindingschema_2_0.xsd"
    version="2.1">
    <globalBindings>
        <serializable uid="1" />
        <xjc:superClass name="the.package.to.my.XmlSuperClass" />
    </globalBindings>
</bindings>