How to add doctype when generating XML from an arbitrary data structure

Question

I was reading SUN's tutorial "Working with XML - PART IV: Using XSLT". Inside this tutorial, in 3, it talks about how to generate XML from an arbitrary data structure making use of a XMLReader and a Transformer. The example only shows adding elements. Is there a way to add DTD information in the XML generated, i.e. <!DOCTYPE ... >? What APIs can I use?
Thanks,
Alex

Answer 1

The simplest way seems to me is to use a XSL file for that. The <xsl:output> attributes doctype-system and doctype-public generate the DTD declaration <!DOCTYPE YOUR_ROOT SYSTEM "yourDTDfile.dtd"> and <!DOCTYPE YOUR_ROOT PUBLIC "yourDTDfile.dtd">, respectively.
When calling transformerInstance.transform() the XSLT processor performs the identity transformation - it just copies elements, attributes, content, processing instructions and comments to the result stream.
If you're using an xsl file for your transformation already, simply add <xsl:output doctype-system="yourDTDfile.dtd"/> to your existing XSL file.
If you're only using the identity transformation you'd need to change the line of code where you obtain the transformer instance from the TransformerFactory to:
t_factory.newTransformer(new StreamSource("test.xsl"));
and use this as test.xsl:
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
   <xsl:output doctype-system="yourDTDfile.dtd"/>
   <!-- this is the identity transformation -->
   <xsl:template match="*|@*|comment()|processing-instruction()|text()">
      <xsl:copy>
         <xsl:apply-templates select="*|@*|comment()|processing-instruction()|text()"/>
      </xsl:copy>
   </xsl:template>
</xsl:stylesheet>Good luck.