Hi Martin,
That's an oversight on my part. Any elements outside of step elements have
to be processed individually. Can I change the for-each-group to just group
adjacent <step> elements? Thank you very much.
Rick
-----Original Message-----
From: Martin Honnen martin(_dot_)honnen(_at_)gmx(_dot_)de
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Saturday, October 26, 2019 1:39 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Feedback on grouping solution
On 26.10.2019 19:03, Rick Quatro rick(_at_)rickquatro(_dot_)com wrote:
I need to process the <step> child elements so that the <figure>
elements are always on the "right" (even-numbered position) in the
output. Immediate children of the <procedure> do not factor into the
odd/even sequence.
The first child of each group of adjacent <step> elements starts a new
odd/even series. To ensure that the each <figure> is in an
even-numbered position, I want to insert a <spacer> element where it is
required.
Here is my stylesheet. My basic question is: is there a better or more
efficient way to do this? I really want to master grouping because it
comes up in a lot of my tasks. Thank you for any input or criticism.
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:xs="http://www.w3.org/2001/XMLSchema";
exclude-result-prefixes="xs" version="2.0">
<xsl:output indent="yes"/>
<xsl:template match="/procedure">
<procedure>
<!-- Group the children of the procedure, keeping
adjacent steps together. -->
<xsl:for-each-group select="child::*"
group-adjacent="local-name(.)">
<xsl:choose>
<xsl:when
test="current-group()[1][not(self::step)]">
<!-- Single element. -->
<xsl:apply-templates
select="current-group()[1]"/>
I don't understand this part, what is supposed to happen with two (or
more) adjacent <note/><note/> or other non step elements? Do you want to
drop any but the first of them?
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