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Re: [xsl] most efficient flat file listing to hierarchical

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David Carlisle wrote:

I started to write an xslt2 soln, but didn't use any xslt2 features, so this
seems to work in 1 as well, it doesn't return the items in quite the
order you asked, but perhaps an xsl:sort would fix that.

Also you didn't say what was the criterion for things being a directory.
this uses the existence of a file in the directory (so empty directories
would be classified as <file/>) an alternative would be to flag
extensionless files as directories (which would make Makefile be classed
as an empty directory, even though it's a file)

David


<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:output indent="yes"/>

 <xsl:template match="listing">
   <xsl:apply-templates select="item"/>
 </xsl:template>

<xsl:template match="item">
<xsl:param name="path" select="''"/>
<xsl:variable name="rel" select="substring-after(.,$path)"/>
<xsl:choose>
<xsl:when test="not(starts-with(.,$path))"/>
<xsl:when test="contains($rel,'/')"/>
<xsl:when test="../item[starts-with(.,concat(current(),'/'))]">
<dir name="{$rel}">
<xsl:apply-templates select="../item">
<xsl:with-param name="path" select="concat(current(),'/')"/>
</xsl:apply-templates>
</dir> </xsl:when>
<xsl:otherwise>
<file name="{$rel}"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>



$ saxon listing.xml listing.xsl
<?xml version="1.0" encoding="utf-8"?>
<dir name="en">
<file name="test.html"/>
<file name="test1.html"/>
<dir name="resource">
<dir name="style">
<file name="test.css"/>
</dir>
</dir>
</dir>
<file name="favicon.ico"/>
<dir name="cn">
<file name="test.xml"/>
</dir>

definately a 'I wish I thought of this' moment....

something like;

              <xsl:sort select="not(contains(.,'.'))"/>
              <xsl:sort order="ascending"/>

should sort things correctly with applied templates as well.

cheers, Jim

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