RE: [xsl] Attributes?

["Michael Kay" <mhk@xxxxxxxxx>]

>     In XSLT 1.0, match="g" matches exactly the same elements as 
>     match="//g".
>     There's a slight difference in 2.0, because match="//g" 
>     will only match g elements that are in a tree with a 
>     document node as its root - which means the processor has 
>     extra checks to do in this case.
> 
> Confused. Don't *all* elements exist in a tree with a document
> node as its root?
> 
> Under what circumstances might an element *not* have such a 
> root please?
> 

No, XSLT 2.0 (and XQuery) allow you to create free-standing nodes. For
example, there was a question a few days ago (I think on the Saxon list
rather than here) from someone who wanted to write a function that returned
a set of attribute nodes:

<xsl:function name="my:atts" as="attribute()">
  <xsl:param name="p"/>
  <xsl:attribute name="a" select="{$p+1}"/>
  <xsl:attribute name="b" select="{$p+2}"/>
</xsl:function>

(Essentially the same as an attribute-set, except that attribute sets do not
allow parameters.)

You can copy these attributes to an element, like this:

<e>
 <xsl:copy-of select="my:atts(17)"/>
</e>

But you can also apply-templates to them:

<xsl:apply-templates select="my:atts(29)"/>

If you try to find the parent of one of these attributes, you find it hasn't
got one:

<xsl:value-of select="count(my:atts(6)/..)"/> displays "0".

And of course the same applies to elements.

Michael Kay