Getting the POSITION() of the PARENT node using Xpath

I just spent waaaaaay too long figuring this out, so I'll leave a note of it for future projects:

To enumerate the current node's position in the current nodeset, position() is great.  However when using nested nodesets, getting the parent's position relative to its nodeset is a bit more complex:

count(parent::*/preceding-sibling::*) + 1 - assuming all nodes in the parent nodeset are of the same type, otherwise:

count(parent::*/preceding-sibling::<type>) + 1 where <type> is the type of the parent node

A quick beakdown:

• parent::* - selects the parent of the current node
• parent::*/preceding-sibling::* - selects the nodes which precede the parent in its nodeset
• count(parent::*/preceding-sibling::*) - counts the nodes preceding the parent
• count(parent::*/preceding-sibling::*) + 1 - if there are n nodes before the parent, then the parent is # n+1

Comments

• Anonymous
  September 07, 2004
  Yeah,
  
  pretty much same as:
  
  count(../preceding-sibling::node()) + 1
  
  Checking for type is reasonable to avoid counting whitespace nodes.
  
  Great post!
• Anonymous
  November 03, 2004
  For the second time in the last month I've bumped into a problem, googled it, clicked on a link to Addy's blog, and found the perfect solution. Getting the POSITION() of the PARENT node using Xpath XmlDocument.CreateProcessingInstruction goes better with...
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