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RE: Accessing node-sets by index with position()

Subject: RE: Accessing node-sets by index with position()
From: "Passin, Tom" <tpassin@xxxxxxxxxxxx>
Date: Tue, 23 Sep 2003 16:44:02 -0400
match position xsl
[Ryan Sawatzky]

> I am trying to iterate through a list of nodes in a node-set by index 
> using the position() function, and I am running into some 
> difficulties.
> 

I have no idea what you want to accomplish or why you are trying to do
it this way.  Please explain what you want to accomplish.  

position() gives you the position of a node within the nodeset being
considered.  However, expressions within the square brackets of a
predicate may behave differently from the way they do outside.  That is
because the predicate gets tested against all possible nodes - it acts
as a filter to select the specified nodes out of all the possible ones.
In your example, there are two possible nodes.  Each one will match
position() for its own position.  No matter how many nodes
$set[position()] might return, you would only ever see the first one
because of the way that value-of works.  So you see '111', the value of
the first one.

You could get the effect you seem to think you want by assigning a
variable to capture the value of position(), like this -

<xsl:for-each select='$set'>
   <xsl:variable name='pos' select='position()'/>
  .... <xsl:value-of select='$set[$pos]'/>
</xsl:for-each>

Although you could do this, I can't make out why you would want to. If
you want to get the next following node, use the following-sibling axis.
Otherwise, you need to explain further what you want.

Cheers,

Tom P


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