Subject: Re: [xsl] Nested for-each-group and current-group() From: "Michael Kay mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Mon, 11 May 2020 16:12:10 -0000 |
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... > ...github.com/wendellpiez... ...gitlab.coko.foundation/wendell...
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