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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: backtracking to find all parents till root[again]
Rahil,
It's still not a watertight problem statement, but the following will produce what you want, and I imagine you can press it into better service if need be. <?xml version="1.0" encoding="iso8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="text" indent="yes"/> <xsl:template match="/">
<xsl:variable name="find" select="'NextTime'"/><xsl:apply-templates select="Top/SubConcepts/SubConcept/Value[contains(., $find)]"/> <xsl:text>Parent: SubConcepts</xsl:text> </xsl:template> <xsl:template match="Value">
<xsl:text>Match Found: </xsl:text><xsl:value-of select="."/>
<xsl:text>
</xsl:text><xsl:apply-templates select="parent::SubConcept[1]"/> </xsl:template> <xsl:template match="SubConcept">
<xsl:text>Parent: </xsl:text><xsl:value-of select="@name"/>
<xsl:text>
</xsl:text><xsl:apply-templates select="preceding-sibling::SubConcept[contains(Value, current()/@name)][1]"/> </xsl:template> </xsl:stylesheet> Regards, --A How do I backtrack to the root element from the result-generating node? Hence if my given structure is of the form: _________________________________________________________________ Express yourself instantly with MSN Messenger! Download today - it's FREE! http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/
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