xsl-list
[Top] [All Lists]

Re: [xsl] Sorting by document order of some other element

2007-06-26 08:36:22
On 6/25/07, David Carlisle <davidc(_at_)nag(_dot_)co(_dot_)uk> wrote:


<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:key name="k" match="set2/*" use="@id"/>
<xsl:output indent="yes"/>
  <xsl:template match="doc">
    <set2>
      <xsl:copy-of select="for $e in set1/* return key('k',$e/@id)"/>
    </set2>
  </xsl:template>
</xsl:stylesheet>


David, that's a cool solution.  Unfortunately I glossed over the fact
that there is some manipulation of the data within the second set
going on.  I'm going to experiment with trying this approach within a
for-each and/or apply-template. as well as attempting to invert the
problem as suggested by Eric.


--
Peter Hunsberger

--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--