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 (5)

  1. Anonymous says:

    Yeah,

    pretty much same as:

    count(../preceding-sibling::node()) + 1

    Checking for type is reasonable to avoid counting whitespace nodes.

    Great post!

  2. Anonymous says:

    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…