Subject: RE: [xsl] how to rearrange nodes based on a dependency graph? From: "Chris Bayes" <chris@xxxxxxxxxxx> Date: Thu, 20 Dec 2001 23:49:31 -0000 |
Ok what about (and I am getting rather blind here) <xsl:template match="/"> <xsl:variable name="allIdsUsed" select="//frag[@requires]"> <xsl:for-each select="@requires"> <ids><xsl:value-of select="." /></ids> </xsl:for-each> </xsl:variable> <xsl:for-each select="xx:node-set($allIdsUsed)/ids"> <xsl:copy-of select="//frag[@id = .]" /> </xsl:for-each> <xsl:copy-of select="//frag[@requires]" /> </xsl:template> This only works if you can get all of the ids out of the @requires idref but like I said I don't use them. If it doesn't then you wil have to split the @requires into the <ids /> somehow. Ciao Chris Forget the bacardi I'm on the red bulls 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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