Re: A little cross referencing problem

Joerg Pietschmann wrote:

> Jeni Tennison <mail@xxxxxxxxxxxxxxxx> wrote:
> > Very sweet :) You could do the same kind of thing for decimal formats
> > to use with format-number().
> > 
> > Of course the only slight drawback is that the document('') call will
> > only get the current stylesheet document, and therefore won't search
> > for keys defined in imported or included stylesheets (or indeed
> > stylesheets in which this one is imported or included).
> > 
> > Perhaps functions like key-available() and decimal-format-available()
> > would be handy.
> 
> Yes. Either this or having something like document(/..) meaning the
> whole compiled style sheet... or, for more fun, document('xsl:this')...
> 
> Really hard-core XSLers will, of course, use recursive templates
> to search all included and imported files. Bonus points for buildung
> a global variable which holds all top-level xsl elements from the
> transitive closure as a node set. :-)

Here's a very simple collection of two templates (not at all hard-core!) that will
merge together all stylesheets (of a stylesheet module) that are referenced from a
main stylesheet directly or indirectly via xsl:include or xsl:import.


<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output method="xml" omit-xml-declaration="yes" indent="yes" />

    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()" />
        </xsl:copy>
    </xsl:template>
    
    <xsl:template match="xsl:include | xsl:import">
       <xsl:apply-templates select="document(@href)/*/node()"/>
    </xsl:template>
</xsl:stylesheet>
 

In the first pass you apply these templates to the root of any stylesheet and get
the output of this transformation into an xsl:variable.

In the second pass (xxx:node-set() will have to be used just once) you convert this
to a node-set and freely search for whatever xsl:key or other elements are of
importance.

Hope this helped.

Cheers,
Dimitre Novatchev.






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