Are you looking for an XSLT 1 solution?
In XSLT 2 or 3, you can do, when in stamp context: <xsl:for-each-group
select="plate" group-adjacent="(@number - 1) idiv 4">, and create a row
for each group.
If you tune the .block width so that 4 of them fit into a .row, you can
also set flex-wrap:wrap in .row, then you need not create div.row at all.
Gerrit
On 28.08.2020 10:44, Mark Wilson mark(_at_)knihtisk(_dot_)org wrote:
This is likely in the archives but I have forgotten how to use them.
Among the other things I have forgotten since I turned 80 is how to a
put set of elements into rows for HTML output. I want to put what four
<plate> elements conjure into each row; the number of< plate> elements
in each <stamp> element varies. The <plate> element references the
identity of a jpeg that will be put on the page. I want four jpegs in
each row.
my css is:
.row {
width: 100%;
display: flex;
flex-direction: row;
justify-content: center;
}
My input XML is a long list of <stamp> elements with <plate> children.
....
<stamp >
<plate number="1"/>
<plate number="2"/>
<plate number="3"/>
<plate number="4"/>
<plate number="5"/>
<plate number="6"/>
<plate number="7"/>
<plate number="8"/>
</stamp>
....
and my XSLT dealing with this within <xsl:template match="stamp"> is:
<div class="row">
<xsl:for-each select="plate">
<div class="block">
<a href="{concat('plates/', @number,
'.jpg')}">
<img class="large"
src="{concat('plates/',@number, 't.jpg')}"/>
</a>
</div>
</xsl:for-each>
</div>
Works fine for four or fewer <plate> elements, but more than that makes
a mess
Any help appreciated,
Mark
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