[xsl] xsl transforming xsl

["Alex Reuter" <xslalex at hotmail dot com>]

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



_________________________________________________________________
Join the world’s largest e-mail service with MSN Hotmail. 
http://www.hotmail.com


XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list