CODEWITHSUNDEEP
javaxmlxsltjaxb
mdm
mdm

Reputation: 3988

JAXB and XSLT processor

I am using JAXB and maven-jaxb2-plugin and I am able right now to bind my schemas to Java code successfully.
I also have a .xsl file "annotate_schemas.xsl" that modifies a specific schema adding some additional information.
Finally, on the schema that I want transformed, I added the header:

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="annotate_schemas.xsl"?>
...

The problem is that, while the .xsl is correct (if I open my schema file in a browser, the transformation is done flawlessly), JAXB ignores it and binds an untouched version of my schema.

My question is: Does JAXB (and/or its plugin) have an XSLT processor?? Is there a way to tell JAXB to bind the result of the XSLT transformation instead of the original?

Thank you very much

Upvotes: 0

Views: 1161

Answers (2)

Michael Kay
Michael Kay

Reputation: 163322

JAXB, like the vast majority of XML-consuming applications, takes no notice of an <?xml-stylesheet?> processing instruction. If you want to transform a document before passing it to JAXB, you need to transform it explicitly, for example by using the JAXP transformation API. (There is an option in JAXP to request transformation according to the value of the xml-stylesheet PI if that's how you want to control it: TransformerFactory.useAssociatedStylesheet()).

Upvotes: 2

Sujay
Sujay

Reputation: 6783

You can try something like this:

TransformerFactory transFact = TransformerFactory.newInstance();
Templates displayTemplate = transFact.newTemplates(new StreamSource(new File("your_xsl_file")));

TransformerHandler handler = 
  ((SAXTransformerFactory) transFact).newTransformerHandler(displayTemplate);

Upvotes: 0

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