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RE: how to rearrange nodes based on a dependency graph?
- From: "Chris Bayes" <chris at bayes dot co dot uk>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Thu, 20 Dec 2001 22:29:52 -0000
- Subject: RE: [xsl] how to rearrange nodes based on a dependency graph?
- Reply-to: xsl-list at lists dot mulberrytech dot com
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
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