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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: xpath - how to return all nodes but the node match
 <xsl:param name="delete_id"/> <xsl:template match="/"> <xsl:copy-of select="//category//artist[@id!=$delete_id]"/> </xsl:template> You select all artist-elements, which don't have the $delete_id. This works for me. But I think you want something different. Try a copy whith each node separately: <xsl:param name="delete_id"/> <xsl:template match="*|text()|@*">
<xsl:copy>
<xsl:apply-templates select="*[@id != $delete_id]|text()|@*"/>
</xsl:copy>
</xsl:template>This stylesheet would copy the complete input to the output. A little change at the apply-templates removes the elements which should be deleted: <xsl:apply-templates select="*[@id != $delete_id]|text()|@*"/> Regards, Joerg XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list 
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