RE: Sibling sort order

David, thanks for your help so far. Based on your information and the
subsequent exchange I think I understand the issue. I certainly understand
it better. If you use a select attribute to obtain a node list,  the
variable element will contain a node list and you can query the path of this
node list using /, // and []. If you do not use a select statement and
instead copy the contents of a node list in the content of the variable, you
obtain a result tree fragment which does not support path based queries.

Therefore, in order to sort the content of the variable, you cannot use a
select statement. Sadly, for my puposes this does not help. The method
behind my madness was yet another grouping methodology. I had intended to
use the relationship between the siblings to determine when the group
field(s) had changed enabling me to process group header(s)/footer(s) etc...
However, in order to do this I need to be able to reference the siblings in
sorted order and I cannot do this unless I use a select statement. It seems
that my only recourse is to use a 2 stage method and transform the data
first into a ordered list before transforming into the output. The example
below illustrates where I am on this one. 

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
    
<xsl:template match="/">
 <xsl:apply-templates/>
</xsl:template>

<xsl:template match="root">  

 <xsl:variable name="x" select="*"/>

 <xsl:variable name="y">
  <xsl:for-each select="*">
    <xsl:sort select="attribute::f1"/>
    <xsl:copy-of select="."/>
  </xsl:for-each>
 </xsl:variable>
 
 <xsl:apply-templates select="$x"/>
 <xsl:apply-templates select="$y"/>
 
</xsl:template>
 
<xsl:template match="row">
 <xsl:variable name="pos" select="position()-1"/>
 <div>
 <xsl:value-of select="attribute::f1"/>
 <xsl:value-of select="parent::*/*[position()=$pos]/attribute::f1"/> 
 </div>
</xsl:template>

</xsl:stylesheet>

<root>
 <row f1="c"/> 
 <row f1="b"/>
 <row f1="a"/> 
</root>


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