On 11.05.2020 17:47, Wendell Piez wapiez@xxxxxxxxxxxxxxx wrote:
Hi,
Alternatively, if you wish to continue using xsl:apply-templates
idiomatically and capture all nodes in traversal without the ungainly
"current-group()/node()", you could write
<xsl:for-each select="current-group()">
<xsl:apply-templates/>
</xsl:for-each>
I think Rick already did a similar thing in the stylesheet that he
posted 4 mins before your message. He introduced two xsl:for-each
instructions (gasp!), thereby totally marring his or rather, his
stylesheetbs HOAXCoQS [1] score, from 26.32 down to 19.05!
[1] https://github.com/sydb/HOAXCoQS
However, note that this will get you a demerit in certain automated
assessments of XSLT code "quality", which disparage any use of
for-each even for simple context switching. (Right Gerrit? ;-)
Thatbs true, but you can compensate it by using xsl:apply-templates
idiomatically, for example. Or another level of nested grouping just for
the score of it.
Grouping is never bad, ask Michael MC<ller-Hillebrand.
Gerrit
Cheers, Wendell
On Mon, May 11, 2020 at 10:44 AM Martin Honnen martin.honnen@xxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
Am 11.05.2020 um 16:34 schrieb Rick Quatro rick@xxxxxxxxxxxxxx:
I am using nested xsl:for-each-group constructs but I don't get the
inner <ul> elements. I think I am incorrect in my use of
current-grouping-key() or current-group() on the inner group. Thanks in
advance.
I think you need to process the child nodes of the current group with e.g.
current-group()/*
in your inner grouping.
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