[xsl] Re: xsl-list Digest 22 Jun 2005 05:10:01 -0000 Issue 455

Subject: [xsl] Re: xsl-list Digest 22 Jun 2005 05:10:01 -0000 Issue 455
From: Michel Charpentier <charpov@xxxxxxx>
Date: Wed, 22 Jun 2005 16:26:15 -0400
On Jun 22, 2005, at 1:10 AM, xsl-list-digest- help@xxxxxxxxxxxxxxxxxxxxxx wrote:

Date: Tue, 21 Jun 2005 23:24:29 +0100
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
From: David Carlisle <davidc@xxxxxxxxx>
Subject: Re: [xsl] Contexts that are atomic values
Message-Id: <200506212224.XAA23479@xxxxxxxxxxxxxxxxx>


Is this behavior normal?


Yes, or at least it will be. (It's best to say if you are using xslt2:-)


distinct-values() as its name suggests returns values not teh nodes the
values they were in, so it's like going


<xsl:for-each select="(1,2,3,4)"

in which the current item would be the integers 1 2 3 4 but being
integers rather than nodes they can't be used as the base for any
relative XPath (or even / as that depends on the current dpcument)

That there's no current node while processing a scalar value, I understand, but why can't there be a current document?



you can go <xsl:variable name="input" select="/"/> as a top level variable (or anywhere outside the for-each) and then use

                 <xsl:apply-templates
                 select="$input/key('by-type',current())"/>
or equivalently

                 <xsl:apply-templates
                 select="key('by-type',current(),$input)"/>

Or perhaps more naturally don't use distinct-values at all, but ciew it
as a grouping problem


<xsl:for-each-group select="item" group-by="@type">
...

for-each-group groups the nodes themselves rather than just the values.

I don't know why I didn't think of using a variable. As for 'for- each-group', I didn't know it existed at all :-) Of, course, the for- each-group approach is the best solution (once you figure out there's also a current-group() function, which is not obvious from the w3.org documentation...).


Thanks a lot for your help,

MC
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