Subject: Re: [xsl] Nested for-each-group and current-group() From: "Imsieke, Gerrit, le-tex gerrit.imsieke@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Mon, 11 May 2020 16:15:40 -0000 |
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.
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