Re: [xsl] Nested for-each-group and current-group()

["Wendell Piez wapiez@xxxxxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

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.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.
>
>
>
> 



-- 
...Wendell Piez... ...wendell -at- nist -dot- gov...
...wendellpiez.com... ...pellucidliterature.org... ...pausepress.org...
...github.com/wendellpiez... ...gitlab.coko.foundation/wendell...