RE: xsl, and different xml versions

the input document ( doc.xml )
=====
<doc>
   <foo/>
   <bar/>
</doc>

   <xsl:import href="doc-a.xsl"/>
   <xsl:import href="doc-b.xsl"/>

   <xsl:template match="/">
       <!--
       Assume we've determined a namespace-uri unique
       to the xsl document to invoke.  This may be an
       attribute of the main xml document, may be passed
       in as a stylesheet parameter, etc.
       -->
       <xsl:variable name="xsldoc-uri" select="'doca-uri'"/>

       <xsl:variable name="selector">
           <xsl:element name="x" namespace="{$xsldoc-uri}"/>
       </xsl:variable>

       <xsl:apply-templates select="xalan:nodeset($selector)//*">
           <xsl:with-param name="root" select="/" />
       </xsl:apply-templates>
   </xsl:template>
</xsl:stylesheet>

   <xsl:template match="*[namespace-uri() = 'doca-uri']">
       <xsl:param name="root" select="/.."/>
       <xsl:apply-templates select="$root" mode="doca-uri"/>
   </xsl:template>

   <xsl:template match="/" mode="doca-uri">
       <doc>
       <xsl:apply-templates mode="doca-uri"/>
       </doc>
   </xsl:template>

   <xsl:template match="foo" mode="doca-uri">
       <foo v="a"/>
   </xsl:template>

   <xsl:template match="bar" mode="doca-uri">
       <bar v="a"/>
   </xsl:template>
</xsl:stylesheet>

   <xsl:template match="*[namespace-uri() = 'docb-uri']">
       <xsl:param name="root" select="/.."/>
       <xsl:apply-templates select="$root" mode="docb-uri"/>
   </xsl:template>

   <xsl:template match="/" mode="docb-uri">
       <doc>
       <xsl:apply-templates mode="docb-uri"/>
       </doc>
   </xsl:template>

   <xsl:template match="foo" mode="docb-uri">
       <foo v="b"/>
   </xsl:template>

   <xsl:template match="bar" mode="docb-uri">
       <bar v="b"/>
   </xsl:template>
</xsl:stylesheet>

<?xml version="1.0" encoding="UTF-8"?>
<doc>
   <foo v="a"/>
   <bar v="a"/>
</doc>

<?xml version="1.0" encoding="UTF-8"?>
<doc>
   <foo v="b"/>
   <bar v="b"/>
</doc>

From: Jake Briggs <jakbri@xxxxxxxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject:  xsl, and different xml versions
Date: Fri, 15 Apr 2005 14:27:34 +1200

Hi all

I have a problem and I am unsure which approach to take in order to solve it. The problem is this:

The root element of the xml I am transforming with xsl will have a version number in it, indicating which version of the xsl i want to transform it with. By that, I mean I have 2 style sheets, both operating on similer xml. I though I may be able to do this by xsl:import, when the xsl matches the root element I though that I could make a decision based on the value of the attribute. IE :

The source xml :

<report version="1">
 <table foo="bar"/>
 <text value="sometext"/>
</report>

The xsl :

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:template match="/">
<xsl:if test="@version = '1'"> <xsl:import href="version1.xsl"/>
</xsl:if>
<xsl:if test="@version = '2'"> <xsl:import href="version2.xsl"/>
</xsl:if>

   <xsl:apply-imports/>
 </xsl:template>

</xsl:stylesheet>


The files version1.xsl, and version2.xsl would contain matches for "table" and "text", but would process them differently.

But no, apparently <xsl:import /> and <xsl:include /> can only have <xsl:stylesheet /> as a parent. I would like to do something like the above, because I dont want to have to call my version 2 xml elements convoluted names so that they dont conflict with the version 1 xml. I dont want to have a version attribute in each xml element either. Also, I would prefer to avoid having the software that initiates the transformation make the decision on which stylesheet to use to transform the xml, but maybe that is the correct way to do it.

Any ideas?

Jake

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