Re: [xsl] What does this XPath expression mean ?

[Kamal <kbhatt@xxxxxxxxx>]

Kevin Grover wrote:
> On Sat, Mar 22, 2008 at 5:43 AM, Kamal <kbhatt@xxxxxxxxx> wrote:
>   
> >  I was in the middle of answering this, and I didn't want my good work to
> >  go to waste, so I thought I would answer again.
> >
> >  Just looking at this:
> >
> >  [not(@label = preceding::*/@label)]
> >
> >
> >  It is comparing the current label is exactly the same as the preceding
> >  labels. For example,
> >
> >  <blah label="A">1</blah>
> >  <blah label="A">2</blah>
> >  <blah label="B">3</blah>
> >  <blah label="B">4</blah>
> >  <blah label="C">5</blah>
> >
> >  This will match on 2, 4, 5. As you can gather, that XPath is pretty much
> >  useless without your dataset being sorted. This is an inefficient way of
> >  grouping values. Your other post about the generate-id is a more
> >  efficient way of doing this. Personally speaking, if you are not worried
> >  too much about efficiency, I recommend sticking with this method. It is
> >  easier to understand and doesn't rely on keys.
> >
> >     
>
> Actually, it picks up 1, 3, and 5 and does not matter if the list is
> sorted (unless you change the relative order of elements with the same
> label).  Why?  When 1 is selected there are no other nodes (preceding
> nodes) with label="A", when 2 is _looked_ at, there is another node
> (preceding node) with label="A" (namely 1), so it's excluded.
>   
Serves me right for answering so late at night :). Apologies, you are 
right about the selection.When I was talking about sorting, I meant 
sorting against label.