Forming the union of two sets that you know to be disjoint is likely to
involve unnecessary sorting and de-duplication operations. Better to use

count(preceding-sibling::*) + count(following-sibling::*)

or why not count(../*)-1 

Michael Kay
http://www.saxonica.com/

> -----Original Message-----
> From: Karl Stubsjoen [mailto:karl@xxxxxxxxxxxxxxxxxxxx] 
> Sent: 15 December 2004 17:47
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject:  Best way to count siblings
> 
> Is this the best way to count siblings?
> 
> <xsl:value-of select="count(preceding-sibling::* | 
> following-sibling::*)"/>