[xquery-talk] distance between elements and hierarchical order

[Charles Brooking]

Wolfgang Hoschek wrote:
> Both can probably be solved by walking the ancestor axis of the 
> elements towards the root.
> For 2) find the nearest common ancestor, and add up the steps.
People can possibly correct me on this, but:

    let $a :=
      element a
      {
        attribute id {"1"},
        element b
        {
          attribute id {"1-1"},
          element c {attribute id {"1-1-1"}},
          element c {attribute id {"1-1-2"}}
        },
        element b
        {
          attribute id {"1-2"},
          element c {attribute id {"1-2-1"}},
          element c {attribute id {"1-2-2"}}
        }
      }
    let
      $x := $a[@id="1"]/b[@id="1-1"]/c[@id="1-1-1"],
      $y := $a[@id="1"]/b[@id="1-2"]/c[@id="1-2-2"]
    let $possible-distances :=
      let
        $x-ancestors := $x/ancestor-or-self::element(),
        $y-ancestors := $y/ancestor-or-self::element()
      for
        $x-ancestor at $x-ancestor-position in $x-ancestors,
        $y-ancestor at $y-ancestor-position in $y-ancestors
      where
        $x-ancestor is $y-ancestor
      return
        count($x-ancestors) - $x-ancestor-position +
        count($y-ancestors) - $y-ancestor-position
    return
      min($possible-distances)

Now a question:

If I had the following as one of the for-clauses:

    $x-ancestor at $x-ancestor-position in $x/ancestor-or-self::element()

would $x itself (always) be at position 1, or should it be last?

Later
Charlie

-- 
Charles Brooking
Associate Software Engineer
Canon Information Systems Research Australia

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