RE: [xsl] How to sort a nodeset referenced by a variable?

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Shailendra,

At 04:19 PM 7/5/01, you wrote:
> I changed the sort key from
> <xsl:sort select="$fileterNode/secondElement">
> to
> <xsl:sort select="./secondelement" />
>
> It started working. I think now I understand your point.

But as has been mentioned two or three times in the last few days,

./secondelement

is just the same (as an XPath expression) as

secondelement

This is because ./secondelement expands to 
self::node()/child::secondelement, and there's no need for the first step, 
since it is just traversing the tree from the context node to itself (and 
only then going to your secondelement children).

So <xsl:sort select="secondelement" /> would work as well (and be a 
teeny-tiny bit faster).

Cheers,
Wendell

At 04:19 PM 7/5/01, you wrote:
> Hi Mike,
> Thanks for the response I got it working.
>
> I changed the sort key from
> <xsl:sort select="$fileterNode/secondElement">
> to
> <xsl:sort select="./secondelement" />
>
> It started working. I think now I understand your point.
>
> regards,
> Shailendra
>
> On Wed, 4 Jul 2001, Michael
> Kay
> wrote:
>
> > > I am doing something like this
> > >
> > > <xsl:variable name="filterdNode" select="//*" />
> > >
> > > <xsl:variable name="sortedList">
> > > <xsl:for-each select="$filterNode/someElement">
> > >  <xsl:sort select="$fileterNode/secondElement">
> >
> > The value of your sort key will be the same for every node in the
> > collection. It's meaningless to use a sort key whose value doesn't 
> depend on
> > the context node.
> >
> > Mike Kay
> > Software AG
> >
> >
> >  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> >
> >
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


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