How to generate classes from XSD that implements serializable?

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Solution 1

If you are using XJC, I recomend you to read this reference: JavaTM Architecture for XML Binding: JAXB RI Vendor Extensions Customizations :

You have to add in your schema aditional namespaces definition to add xjc aditional markup:

<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"

           xmlns:jaxb="http://java.sun.com/xml/ns/jaxb"
           xmlns:xjc="http://java.sun.com/xml/ns/jaxb/xjc"
           jaxb:extensionBindingPrefixes="xjc"
           jaxb:version="1.0">

Then, including an <xjc:serializable> node within <jaxb:globalBindings>:

<xs:annotation>
   <xs:appinfo>
      <jaxb:globalBindings generateIsSetMethod="true">
          <xjc:serializable uid="12343"/>
      </jaxb:globalBindings>
   </xs:appinfo>
</xs:annotation>

This will cause that all the concrete classes implement the Serializable interface. Also, you can define the UUID value of the resulting classes (that's an optional attribute).

Solution 2

I've found

<schema
  xmlns="http://www.w3.org/2001/XMLSchema"
  xmlns:jaxb="http://java.sun.com/xml/ns/jaxb"
  xmlns:xjc="http://java.sun.com/xml/ns/jaxb/xjc"
  jaxb:extensionBindingPrefixes="xjc"
  jaxb:version="1.0"  
  >

  <!-- FORCE ALL CLASSES IMPLEMENTS SERIALIZABLE -->
  <annotation>
    <appinfo>
      <jaxb:globalBindings generateIsSetMethod="true">
        <xjc:serializable uid="1"/>
      </jaxb:globalBindings>
    </appinfo>
  </annotation>

   ....

</schema>
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Topera

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Updated on September 26, 2020

Comments

  • Topera
    Topera over 3 years

    I need to generate many classes from my XML Schema (XSD) in a package (.jar). How can I configure these classes to be serializable?

    (I'm using Eclipse and JAX-B)