Re: [xsl] xsl:for-each-group and identity transform

[David Carlisle <davidc@xxxxxxxxx>]

> I guess is never going to work, we need "current-group()". I simply
> can't see how we can get the modified identity template into play.

not in the templates that you use to process the group (at least not
normally). I had to make some guesses as I had no input or expected
output but as I say the usual model is that you just have

<xsl:apply-templates select="current-group()"/>

as the content of the for-each-group, perhaps with a mode, if that's
needed, and then template matching proceeds as normal and can be the
identity template or any other.

<xsl:copy-of select="./node()"/>


or just node() as ./ doesn't do anything.

David

________________________________________________________________________
The Numerical Algorithms Group Ltd is a company registered in England
and Wales with company number 1249803. The registered office is:
Wilkinson House, Jordan Hill Road, Oxford OX2 8DR, United Kingdom.

This e-mail has been scanned for all viruses by Star. The service is
powered by MessageLabs. 
________________________________________________________________________