Subject: Re: [xsl] Transforming tabular information to hierarchical From: "Andrew Welch" <andrew.j.welch@xxxxxxxxx> Date: Tue, 13 Feb 2007 16:59:15 +0000 |
If I have a tabular data set that defines parent-child relationships, is it possible to transform this into a hierarchical tree format using XSLT?
Here's a csv to XML converter I wrote the other day, feel free to convert it to use tabs, or adjust your input to be comma delimited.
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" exclude-result-prefixes="xs">
<xsl:template name="main"> <xsl:choose> <xsl:when test="unparsed-text-available($pathToCSV)"> <xsl:variable name="csv" select="unparsed-text($pathToCSV)"/> <xsl:variable name="lines" select="tokenize($csv, '
')" as="xs:string+"/> <xsl:variable name="elemNames" select="tokenize($lines[1], ',')" as="xs:string+"/> <root> <xsl:for-each select="$lines[position() > 1]"> <row> <xsl:variable name="lineItems" select="tokenize(normalize-space(.), ',')" as="xs:string+"/> <xsl:for-each select="$elemNames"> <xsl:variable name="pos" select="position()"/> <xsl:element name="{normalize-space(.)}"> <xsl:value-of select="$lineItems[$pos]"/> </xsl:element> </xsl:for-each> </row> </xsl:for-each> </root> </xsl:when> <xsl:otherwise> <xsl:text>Cannot locate : </xsl:text><xsl:value-of select="$pathToCSV"/> </xsl:otherwise> </xsl:choose> </xsl:template> </xsl:stylesheet>
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