How to perform set operations in XPath 1.0

Question

I've seen on SO and other places that the following is supposed to work (this example is lifted directly from O'Reilly's XSLT Cookbook):

(: intersection :)
$set1[count(. | $set2) = count($set2)]

(: difference :)
$set1[count(. | $set2) != count($set2)]

and it looks like it should be OK, however this seems to fail when used with actual paths rather than variables. For example, given the following document

<a>
  <new>
    <val>1</val>
    <val>2</val>
  </new>
  <old>
    <val>2</val>
    <val>3</val>
  </old>
</a>

and the XPath expression /a/new/val[count(. | /a/old/val)=count(/a/old/val)]/text() I would expect to get the node-set { 2 } but instead am getting { 1 2 }. Any ideas what I'm doing wrong?

Good question, +1. See my answer for explanation of your problem and for a simple and easy one-liner XPath expression - solution -- no other answer proposes a solution at this moment. :) — Dimitre Novatchev, Aug 25 '11 at 04:21

score 11 · Accepted Answer · answered Aug 25 '11 at 04:18

11

The formulas for node-set intersection use node-identity, not value identity.

Two nodes are identical if and only if count($n1|$n2) =1

However, you want to intersect based on value identity.

Solution:

Use:

/a/new/val[. = /a/old/val]

this selects any /a/new/val for which there exists at least one /a/old/val element such that the string values of these two elements is the same.

answered Aug 25 '11 at 04:18

Dimitre Novatchev

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Nice answer! Can we do implement set *Minus* operations also ? – Arup Rakshit Sep 09 '13 at 18:09
Ok... Set *Minus* is `/a/new/val[not(. = /a/old/val)]`. Am I right? – Arup Rakshit Sep 09 '13 at 18:31
@Babai, Yes, but see the statement that in XPath set means something different. – Dimitre Novatchev Sep 09 '13 at 19:42
I didn't catch you.. could you repeat please ? – Arup Rakshit Sep 09 '13 at 19:43
@Babai, "The formulas for node-set intersection use node-identity, not value identity" – Dimitre Novatchev Sep 09 '13 at 20:18
Okay.. Got now. But then what about *Minus* operations? – Arup Rakshit Sep 09 '13 at 20:19
1

@Babai, If you don't mix node identity with value identity and always remember that these define two different concepts of "sets", then this is OK. – Dimitre Novatchev Sep 10 '13 at 02:23
unexpectedly powerful operator! it can implement the function of "in" in SQL, but it seemed that no formal specification about it can be found! – frankli22586 Nov 25 '14 at 08:15
@frankli22586, Yes, this was discovered by Michael Kay. In XPath 2.0 there is the `intersect` operator, which removes the need for tricks and gives way for efficient implementations. – Dimitre Novatchev Nov 25 '14 at 14:36

Emiliano Poggi · Answer 2 · 2011-08-24T17:22:23.057

Remark that intersection always results in a single node-set composed by the nodes in common between the two original nodes-sets.

Also note that two nodes with same name and content are to be treated as two distinct nodes. Thus /a/new/val/text() and /a/old/val/text() have same value but they are completely distinct text nodes.

Thus your current intersection:

/a/new/val[count(. | /a/old/val)=count(/a/old/val)]

should evaluate to an empty node-set, because you are intersecting two nodes-sets without any node in common (count() operation will never match). You are doing something like this:

/a/new/val ∩ /a/old/val = ∅

While /a/new ∩ /a/old/preceding::new will produce new.

How to perform set operations in XPath 1.0

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