Re: [xsl] xsl:for or xsl:repeat

["Dimitre Novatchev" <dnovatchev@xxxxxxxxx>]

>  for example you can use the stylesheet
> document('')//node()
> and have as many nodes as you want in that (add 1000 empty comments at
> the end if need be)

Usually a much bigger number can be obtained by:


  document('')//node() | document('')//namespace::* | document('')//@*


because any namespace node is propagated down to all descendants of
the element in which it initially was specified.


So the total number of namespace nodes is roughly O(N^2), where N is
the average depth of the stylesheet's xml tree.


Cheers,
Dimitre Novatchev


On 5/24/06, David Carlisle <davidc@xxxxxxxxx> wrote:
> > Just a quick, somewhat stupid question that probably won't come up in
> > any real practicality.  What if there is the need to iterate more than
> > there are nodes in the document?
>
> well for any fixed number you can ensure that the document you use for
> the iteration (which doesn't have to be your source document) has
> enough nodes: for example you can use the stylesheet
> document('')//node()
> and have as many nodes as you want in that (add 1000 empty comments at
> the end if need be)
> if you are picking up a number dynamically it's probably easier to just
> give in and write the recursive template (which after all is only 4 or 5
> lines of code) apart from anything else loading a massive document just
> tp iterate overthe integers is a bit expensive.
> But if you really want to use //node() then
> z=document('')//node())
> m=count($z)
> a=floor($n div $m)
> b=$n - $a*$m
> for-each $z[position() <= $a]
>  for-each $z
> for-each $z[position() <= $b]
>
> will do any number of iterations up to the square of the number of nodes
> in the stylesheet which can be a pretty big number.
> (I may have got the sums wrong but too late to think about that too
> hard)
>
> David
>
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--
Cheers,
Dimitre Novatchev
---------------------------------------
Truly great madness cannot be achieved without significant intelligence.