This looks like a nested group-starting-with / group-adjacent to me at first
glance.
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-----Original Message-----
From: "Michael Müller-Hillebrand mmh(_at_)docufy(_dot_)de"
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com>
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Sent: Sun, 03 May 2020 10:33
Subject: Re: [xsl] Joining list fragments (was: Handling position in non-atomic
sequences)
Am 02.05.2020 um 13:43 schrieb Michael Kay:
You're talking about a solution, but I don't think you've told us what the
problem is. What are you trying to achieve?
Fair enough,
sometimes one is so deep wrapping one’s thoughts around a detail problem, that
the broader picture gets lost.
We want to join list fragments and some content in between them. An HTML-ish
version of the input looks like this:
<div>
<h2 id="E2">Item with content to be joined follows div to
collect</h2>
<div>
<ol data-meta="listlevel=start">
<li>
<p>1st item</p>
</li>
</ol>
<div class="box" data-meta="collect">
<p>Hint</p>
</div>
<ol data-meta="listlevel=continue">
<li data-meta="listitem=continue">
<p>Para ff</p>
</li>
<li>
<p>2nd item</p>
</li>
</ol>
<p>Other arbitrary content</p>
</div>
</div>
Every broken list sequence starts with data-meta="listlevel=start" and a list
or a list item that is supposed to be joined with the start list is marked
using data-meta="listlevel=continue" and data-meta="listitem=continue". There
can be any number of collect items between lists and multiple continue lists,
but it is guaranteed that whatever needs to be collected will end with a list.
In DTD content model notation: startList, (collectItem*, continueList)+
The lists are not limited to a single level. Gladly, if there is a
"listitem=continue" in a continue list, it is guaranteed to be at the same
level the previous list ends.
The task is to add to the last item of the previous list:
* all content marked "collect" between the lists; other content would break the
process
* content of the next list’s first list item if marked "listitem=continue"
The remaining content of each continue list would be added as additional items
to the start list.
The desired result for the input data above would look like this:
<div>
<h2 id="E2">Item with content to be joined follows div to
collect</h2>
<div>
<ol data-meta="listlevel=start">
<li>
<p>1st item</p>
<div class="box" data-meta="collect">
<p>Hint</p>
</div>
<p>Para ff</p>
</li>
<li>
<p>2nd item</p>
</li>
</ol>
<p>Other arbitrary content</p>
</div>
</div>
All the "collect" content and continue lists are at the same hierarchical level
as the start list. So my initial strategy was to begin with the start list and
collect all "valid" siblings (collect, listlevel="continue") by walking the
following-sibling axis one by one. The result would be a sequence of elements
that have to be processed into the start list at various locations.
My original question was how to find the position of the next <ol> in that
sequence to easily use subsequence(). And I got some helpful pointers for that
(I even looked at xsltfunctions.com <http://xsltfunctions.com/>, but due to my
xsl:copy-of decision node identity seemed not to be an option), thanks a lot!
But currently I am doubting my general strategy and I have the feeling I am
missing one very obvious thing. The task is basic tree transformation, and my
current ideas all look very complicated. For the simplified data I cannot yet
share XSLT code.
Basically I use a mode to process the continue lists and their items. Within
each template I add special handling for "continue" content (ignoring the
elements, just processing their content). For the very last item in each list I
add rules to process the collect elements. Getting to the content of a <li
data-meta="listitem=continue"> and making sure it is processed only once is
where I am stuck at.
Currently I think of putting more effort in the collect phase, e.g. to already
split continue lists in two parts: the continue item, which is then easily
processed with the collect items, and the rest of the list, which will always
be new items or sub-lists.
Pointers if anyone has had success for a problem like this, would be very
welcome. (And I wonder if I have to deal with position in element sequences at
all.)
- Michael Müller-Hillebrand
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