Subject: Re: [xsl] Find elements whose ancestors are the same From: JBryant@xxxxxxxxx Date: Mon, 26 Sep 2005 16:45:58 -0500 |
As a potential next step from Wendell's reply, you can do multi-pass processing in a single stylesheet by using modes and variables. Once you get it working in two stylesheets, come ask us how to get it done in one stylesheet. Jay Bryant Bryant Communication Services (presently consulting at Synergistic Solution Technologies) Wendell Piez <wapiez@xxxxxxxxxxxxxxxx> 09/26/2005 04:00 PM Please respond to xsl-list@xxxxxxxxxxxxxxxxxxxxxx To xsl-list@xxxxxxxxxxxxxxxxxxxxxx cc Subject Re: [xsl] Find elements whose ancestors are the same Ted, In XSLT 2.0, this is much easier with a key: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0"> <xsl:key name="folder-by-ancestry" match="folder" use="string-join(ancestor-or-self::*/@name,'#')"/> <xsl:template match="folder"> <xsl:copy-of select="key('folder-by-ancestry',string-join(ancestor-or-self::*/@name,'#'))"/> </xsl:template> </xsl:stylesheet> (# is used as a delimiter on the assumption it's not allowed inside names.) Unfortunately, XSLT/XPath 1.0 give no function that allows you to stitch together all the names you need into a single key value. Accordingly I'd probably use two passes to do this in 1.0: in the first pass, annotate all the elements with a code (created by walking through the tree and collecting @name attributes); in the second, retrieve elements with the same code. Pass 1: <xsl:template match="folder"> <xsl:copy> <xsl:copy-of select="@*"/> <xsl:attribute name="ancestry"> <xsl:for-each select="ancestor-or-self::*/@name"> <xsl:value-of select="."/> <xsl:text>#</xsl:text> </xsl:for-each> </xsl:attribute> </xsl:copy> </xsl:template> Pass 2 <xsl:key name="folder-by-ancestry" match="folder" use="@ancestry"/> <xsl:template match="folder"> <xsl:variable name="ancestry"> <xsl:for-each select="ancestor-or-self::*/@name"> <xsl:value-of select="."/> </xsl:for-each> </xsl:variable> <xsl:copy-of select="key('folder-by-ancestry',$ancestry)"/> </xsl:template> Cheers, Wendell I'd At 04:38 PM 9/26/2005, you wrote: >Thanks again to everyone who has replied to these messages. I had no >idea how hard this would be. It seemed so simple when I explained it >to the client ;-) > >I'm looking for an XPath expression that will find all elements >whose ancestors are the same. By "same" I mean their @name is the >same at each level of ancestry (and they appear in the same order). > >For example, this expression would return a count of 2 for the >following XML when filtered on the "daughter" folder (or "mother" or >"grandfather" for that matter). > ><root> > <folder name="grandfather"> > <folder name="mother"> > <folder name="daughter" /> > </folder> > </folder> > <folder name="grandfather"> > <folder name="mother"> > <folder name="daughter" /> > </folder> > </folder> > <folder name="grandmother"> > <folder name="father"> > <folder name="son" /> > </folder> > </folder> ></root> > >I know it's going to be something like this, but I just can't get my >head around it! > ><xsl:copy-of select="/root//folder[@name = >following-sibling::*/@name and ancestor-or-self::*/@name = >following-sibling::*/@name/ancestor-or-self::*/@name" /> ====================================================================== Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx Mulberry Technologies, Inc. http://www.mulberrytech.com 17 West Jefferson Street Direct Phone: 301/315-9635 Suite 207 Phone: 301/315-9631 Rockville, MD 20850 Fax: 301/315-8285 ---------------------------------------------------------------------- Mulberry Technologies: A Consultancy Specializing in SGML and XML ======================================================================
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