Re: [xsl] Select nodes with equal position

Subject: Re: [xsl] Select nodes with equal position
From: David Carlisle <davidc@xxxxxxxxx>
Date: Sun, 3 Apr 2005 13:29:47 +0100
  Thank you David. This also works -
  <xsl:value-of select="../../colspec[count(current()/preceding-sibling::entry)+1]/@colname"/>
  (got idea from your answer)

  Seems more programmer friendly to me(not a XPath expert like you)   

The system most likely already knows position() but counting preceding
siblings can be an expensive operation, although actually here, for table
columns, we are presumably talking about 4 or 5 siblings not 4 or 5
thousand, so efficiency may not be a concern.

There is an alternative of course which doesn't require counting at all,
instead of iterating through one set of nodes and then indexing into the
other set using this number, you can iterate over both sets in parallel
passing each node into a template as a parameter and iterating by
selecting the next following sibling of _each_ node.


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