Re: [xsl] Find elements whose ancestors are the same

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
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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
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