Re: [xsl] Trying to undestand why template works

Subject: Re: [xsl] Trying to undestand why template works
From: "Jon Gorman" <jonathan.gorman@xxxxxxxxx>
Date: Thu, 23 Feb 2006 12:31:43 -0600
On 2/23/06, yguaba@xxxxxxxxxxxx <yguaba@xxxxxxxxxxxx> wrote:
> Hello,
> After some digging and experimenting, I finally got a stylesheet to
> do what I wanted. Among other things, it uses the template below.
> Problem is, I don't really understand WHY this template works, and
> would very much like to.

Ooooohhhh boy.  The random code messing to what might work.

> Doesn't "position() > 1" evaluate to either 0 or 1?

Errr, no.  It would either be true or false.  ie

select="foo[position() > 1]"

can be read as "get all the children element of the current node that
are named foo and also have a position greater than one".

Since you are adding them in a sequence, you are adding the one with
position 1 to the result of the recursive call, so you only want those
that are after position 1.

I'm not sure if you are confusing the [] with some sort of array call.
 It's a conditional, so using foo[1] is really short for
foo[position() = 1], which could actually be expanded more as well.

The other thing that might be confusing you is how you're manipulating
the node-set.

Say you have something like


The first time you call this function, you're passing in the node-set
(called node, a bit confusing).  It contains three nodes, all of them

Then the recursive function finds the value of the first el +
sum-value(second el, third el)

this is because both the second el and third el have positions greater
than 2, so they are put into a new node-set and passed along with the
$node param.

second pass
value of second el + sum-value (third el)
In the new node set, only the third el has a position greater than one

third pass

value of third + sum-value ()
now the node-set only has the third el, which has a position of 1

fourth pass
base case, return 0
empty node-set, return 0

0 + 3

3 + 2

5 + 1


Does any of this help?

Jon Gorman

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