Subject: RE: [xsl] ID refs From: Jarno.Elovirta@xxxxxxxxx Date: Mon, 19 Jan 2004 14:36:21 +0200 |
Hi, > I have the following XML: > > > <Specification_expression id="specification_expression-1"> > <id>teste</id> > <description>teste</description> > <operation>or_operator</operation> > <operand> > <sos ref="sos-1"/> > </operand> > </Specification_expression> > > <sos id="sos-1"> > <Specification ref="specification-7"/> > </sos> I assume these two XML fragments exist in the same document. If so, <xsl:template match="sos[@ref]"> <xsl:copy> <xsl:apply-templates select="//sos[@id = current()/@ref]/node()"/> </xsl:copy> </xsl:template> <xsl:template match="sos[@id]"/> <xsl:template match="@* | node()"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> </xsl:copy> </xsl:template> Will get you there. If you have a DTD that defines the "id" attribute as an ID, then <xsl:apply-templates select="id(@ref)/node()"/> Will also get you there-again, there are other ways, too. Cheers, Jarno - This Morn' Omina: One Eyed Man XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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