Altova Mailing List Archives>Archive Index >xsl-list Archive Home >Recent entries >Thread Prev - [xsl] XSL processor version detection stylesheet with XSLT 2.0 properties >Thread Next - [xsl] XSL Help Re: [xsl] Reading recursive from a listTo: Date: 10/1/2007 1:21:00 PM Houman Khorasani wrote: My Selection list: <UnitInfos xmlns="Animals"> <Type>UNIT_LION</Type> <Type>UNIT_DOG</Type> </UnitInfos> According to my selection List, I would like to copy all UnitInfo elements with their children that have a type element which is listed in the small list above. Here is an XSLT 1.0 stylesheet that has the "selection list" in the stylesheet: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:a="Animals" exclude-result-prefixes="a" version="1.0"> <xsl:output method="xml" indent="yes"/> <xsl:strip-space elements="*"/> <UnitInfos xmlns="Animals"> <Type>UNIT_LION</Type> <Type>UNIT_DOG</Type> </UnitInfos> <xsl:template match="@* | node()"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> </xsl:copy> </xsl:template> <xsl:template match="a:UnitInfos"> <xsl:copy> <xsl:apply-templates select="@*"/> <xsl:apply-templates select="a:UnitInfo[a:Type = document('')/xsl:stylesheet/a:UnitInfos/a:Type]"/> </xsl:copy> </xsl:template> </xsl:stylesheet> -- Martin Honnen http://JavaScript.FAQTs.com/ | ||||||
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