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Subject: xsl transforming xsl
From: "Alex Reuter" <xslalex@xxxxxxxxxxx>
Date: Thu, 19 Sep 2002 16:55:26 +0000


Hi List,
I am trying to write an xsl which takes in an xsl template file as input, parses that file, and outputs an initial call-template node for use in building an xsl page for the application.


So the template would be something like:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template name="link">
<xsl:param name="linkid"/>
<xsl:param name="linkname"/>
<xsl:param name="pagename"/>
<xsl:param name="class"/>


</xsl:template>
</xsl:stylesheet>
And I would like the output to be something like:
<xsl:call-template name="link">
<xsl:with-param name="linkid"></xsl:with-param>
<xsl:with-param name="linkname"></xsl:with-param>
<xsl:with-param name="pagename"></xsl:with-param>
<xsl:with-param name="class"></xsl:with-param>
</xsl:call-template>


Ok, heres the problem I am having:
The xsl I am writing needs to reference the xsl namespace in order to properly access the xsl nodes within the template using XPath. But if I do not declare the xmlns:xsl="http://www.w3.org/1999/XSL/Transform", the XPath does not work. If I do declare the xmlns:xsl="http://www.w3.org/1999/XSL/Transform" then outputting the xsl:templates doesn't work because the processor tries to execute the xsl rather than just outputting it.


I cannot output an arbitratry namespace for the xsl template which is generated and use namespace alias for mainly cosmetic reasons, but I think management will not bend on this.

Does anyone have an idea of what might work in this situation? Is there a way to switch the namespace prefix as well as the definition of that namespace?

Thanks,
Alex



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