Re: [xsl] First attempt at xsl:result-document

Hi, Spencer,

When I tried your XML and XSL files (after filling in the necessary 
pieces), I found that I got no output unless I removed the "file:///" part 
of the href value.

Jay Bryant
Bryant Communication Services
(presently consulting at Synergistic Solution Technologies)





Spencer Tickner <spencertickner@xxxxxxxxx> 
04/25/2005 11:59 AM
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Subject
[xsl] First attempt at xsl:result-document






Hi everyone, and thanks in advance for the help.

I have a magical and wonderful xsl that was doing everything I needed
it too. Unfortunately as this buisness goes, requirements changed.
Some of the html documents I was producing were getting too large for
our website. The decision was made to split the larger ones up by
part. So after some research I found xsl:result-document. Here is some
sample xml, my single file xsl transformer and my attempt at
multi-file transformation.

xml

<act>
<part>this is a part we will divide a file on</part>
<section>This is a section</section>
<clause>This is a clause</clause>
<part>This is another part, in the new scheme of things, a second 
file</part>
<section>Yet another section</section>
</act>


original xsl (works fine)

<xsl:template match="act">

<html>
<body>
<xsl:apply-templates mode="tableofcontents"/>
<xsl:apply-templates mode="content"/>
</body>
</html>

</xsl:template>
<!-- Down here of course I have the templates that apply the styles
for either mode -->


new xsl (well, not so fine)

<xsl:template match="act">

<xsl:result-document href="file:///toc.html" format="html">
<html>
<body>
<xsl:apply-templates mode="tableofcontents"/>
</body>
</html>
</xsl:result-document>

<xsl:for-each select="part">
 <xsl:variable name="filename" select="concat('file:///', position(), 
'.htm'"/>
<xsl:result-document href={$filename}" format="html">
<html>
<body>
<xsl:apply-templates mode="content"/>
</body>
</html>
</xsl:result-document>
</xsl:for-each>
</xsl:template>

<!-- Exact same  templates that apply the styles for either mode as
original xsl -->


In the new xsl, I get the tableof contents no problem. in terms of
content I get a number of files (same as the number of parts) with no
content in them. I realize that the for-each statement probably
doesn't do what I'm hoping it will do, but I can't quite wrap my mind
around any other ways of doing this. I would really appreciate any
suggestions or advice.

Thank you all very much,

Spencer

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