Xsl:include, dynamic href attribute

Question

Is it possible to use a xsl parameter or variable for the href attribute of xsl:include ?
I tried
<xsl:param name="xxx" />
and then <xsl:include href="{$xxx}"/>
or even
<xsl:include>
<xsl:attribute name="href">
<xsl:value-of select="$xxx" />
</xsl:attribute>
</xsl:include>
None of this works (I haven't tried with other xslt processors than Oracle).
Any suggestions ?
null

Answer 1

I have attempted the same thing before as well. The only way I was able to dynamically build the xsl:include was to declare a new namespace e.g xmlns:out="output.xsl" in my XSL file and do something like the following:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:out="output.xsl">
<xsl:namespace-alias stylesheet-prefix="out" result-prefix="xsl"/>
<xsl:varible name="xxx" select=" 'myFile.xsl' "/>
<xsl:template match="/">
<out:stylesheet version="1.0">
<out:template match="/">
<out:include>
<xsl:attribute name="href">
<xsl:value-of select="$xxx" />
</xsl:attribute>
</out:include>
</out:template>
</out:stylesheet>
</xsl:template>
</xsl:stylesheet>
When applied to any valid XML document this produces the XSL file you are after as the output.
To use this output against your intended XML source consider the following: A Java program can very quickly apply the above stylesheet to any XML source to get the output you want... then apply that XSL result to the actual XML document you are wanting to transform. From here you can do what you want with the transformed result.