Subject: Re: [xsl] most efficient flat file listing to hierarchical From: James Fuller <jim.fuller@xxxxxxxxxxxxxx> Date: Thu, 11 Jan 2007 13:54:10 +0100 |
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>
<xsl:sort select="not(contains(.,'.'))"/> <xsl:sort order="ascending"/>
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