Re: Nested for-each-group and current-group()

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@xxxxxxxxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> 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.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...
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