Looking at the code, I assume what makes the difference is that without
xsl:for-each-group, all "b" and "c" elements within their parent element
would go into the same container1 instance instead of each "b"/"c"
sequence receiving their own container. Am I right?
the usage of for-each-group that you suggested, with a constant grouping
key, is exactl_ the same as the version I posted that did not use
for-each-group at all. In both cases they will only make a single
container1 element and place into that all of the relevant elements even
if they occur later in the input sequence.
the "two methods" that i posted in the earlier reply were designed to
handle this case, they will separately group each contiguous run of b,c
and d,e,f groups into possibly multiple container element_s_.
If so, it won't hurt me, as my "b" and "c" elements will always appear
before any "d" element, therefore I think can do without
xsl:for-each-group.
so, you don't need for-each groop here.
The lesson I learned: The hardest part with XSLT (2.0) is to find its
easy solution to your complex problem. ;-)
sounds about right!
David
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