If the tool gives you downvotes for using xsl:for-each, just switch to
xsl:iterate instead. It won't improve your code, but it will earn you brownie
points. These tools are designed to create perverse incentives, aren't they??
Michael Kay
Saxonica
On 11 May 2020, at 16:47, Wendell Piez wapiez(_at_)wendellpiez(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> 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>
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? ;-)
Cheers, Wendell
On Mon, May 11, 2020 at 10:44 AM Martin Honnen
martin(_dot_)honnen(_at_)gmx(_dot_)de
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Am 11.05.2020 um 16:34 schrieb Rick Quatro rick(_at_)rickquatro(_dot_)com:
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.
--
...Wendell Piez... ...wendell -at- nist -dot- gov...
...wendellpiez.com... ...pellucidliterature.org... ...pausepress.org...
...github.com/wendellpiez... ...gitlab.coko.foundation/wendell...
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