On Fri, Jan 31, 2014 at 5:36 PM, Dimitre Novatchev
<dnovatchev(_at_)gmail(_dot_)com> wrote:
On Fri, Jan 31, 2014 at 9:16 AM, Wendell Piez
<wapiez(_at_)wendellpiez(_dot_)com> wrote:
For this, one could also say
a[@href = (../a except .)/@href]
Hi Wendell,
I agree with every statement in your message.
At the same time, I believe, that the OP wouldn't have posted the
question in case he was allowed to use XSLT 2 -- as it is obvious that
in XSLT 2 one should use the <xsl:for-each-group> instruction.
In the unlikely case that someone is using XSLT 2 and still trying to
use keys for grouping, this most probably means that such person isn't
aware of the existence of the <xsl:for-each-group> instruction.
http://www.biglist.com/lists/lists.mulberrytech.com/xsl-list/archives/201211/msg00170.html
OP is using XSLT 2. OP only wants to find what things have duplicates.
Thats not a grouping problem, it's a problem that a certain variant of
for-each-group happens to solve but it's not the simplest way of
solving it.
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--