Re: [xsl] position() returns what?

[David Carlisle <davidc@xxxxxxxxx>]

> But this outputs equal numbers: 2,4,6,8,10,...
> I did expect that it would output 0,1,2,3,4,5

This is a FAQ (I'm sure itmustbe in teh faq file for this list)
position() returns a number that is not a property of teh node (teh same
node can have any number of position() values, for example if you select
it with .. or . it always has position() =1.

position() returns the position of the node in the current node list,
and if that is the default selection for apply-templates this is
child::node(0 which is all nodes including text nodes, so
in your case the odd numbered nodes are all white space text nodes, if
you added a template for them,
<xsl:template match="text()">
     <xsl:value-of select="position()"/>
then you'd see it.

change you apply tempaltes to use select="*" if you only want to process
element children, or use xsl:strip-space to discard teh white space
before processing begins.

David

-- 
http://www.dcarlisle.demon.co.uk/matthew

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