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RE: [xsl] how to rearrange nodes based on a dependency graph?
Subject: RE: [xsl] how to rearrange nodes based on a dependency graph? From: "Chris Bayes" <chris@xxxxxxxxxxx> Date: Thu, 20 Dec 2001 22:29:52 -0000 |
Or even <xsl:key name='fragkey' match='frag' use='@id' /> <xsl:template match="frag[@requires]"> <xsl:for-each select="@requires"> <xsl:for-each select="key('fragkey',.)[not(@requires)]"> <xsl:copy-of select="." /> </xsl:for-each> </xsl:for-each> <xsl:copy-of select="." /> </xsl:template> <xsl:template match="frag" /> Hey it's christmas. I don't use ids and idrefs. If this <xsl:for-each select="@requires"> returns all of the idrefs then this might work. But if it doesn't then mine is a stella and a bacardi and coke hick! Ciao Chris XML/XSL Portal http://www.bayes.co.uk/xml XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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