Re: Balanced Two column tables

[David Carlisle <davidc@xxxxxxxxx>]

> Yes. Where can I find some documentation on node-set()? I'm using 
> XT to do the processing.

on xt's html documentation pages, although probably more useful to scan
the archives of this list for examples.

> Any suggestions on how to apply it to my current problem gratefully 
> received.

well if before you had

<xsl:apply-templates select="asdlkjfhba[ajshgc]//*[lkagdc]"/>

and now you want to apply the same templates but first sort the list
into order, do

<xsl:variable name="x">
 <xsl:for-each select="asdlkjfhba[ajshgc]//*[lkagdc]">
   <xsl:sort select="jhfvgkjb"/>
   <xsl:copy-of select="."/>
 </xsl:for-each>
</xsl:variable>

so now you have your nodes in sorted order except they are rather
unhelpfully in a result-tree-fragment not a node set so you can't access
them, but not being able to access them is more or less the only
difference between a node set and a result tree fragment (the other
being the method of coercion to booleans) and so xt and saxon provide
a node-set fuction and microsoft rather unhelpfully provide implicit
coercion that gives you back a node-set.

so to apply your templates you now do

<xsl:apply-templates select="xt:node-set($x)/*"/>


xt:node-set() works like document() you get a new input tree with its
own root and all teh axes like following-sibling work on the document
order of _that_ tree (ie sorted order) not on the document order in
the original document.

Note that even though you get a node-set and not a node list, the order
is preserved, the node set just has _one_ node, the root node, and
the nodes you are interested in are in sorted order.

That's the reason for the /* in the xt:node-set($x)/* to make sure you
get the nodes you want rather than the root node, which may or may not
give you the same thing, depending on what your template matching "/"
does.



David


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