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(_at_)mulberrytech(_dot_)com>
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[(_at_)name =
following-sibling::*/@name and ancestor-or-self::*/@name =
following-sibling::*/@name/ancestor-or-self::*/@name" />
======================================================================
Wendell Piez
mailto:wapiez(_at_)mulberrytech(_dot_)com
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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