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Re: [xsl] XPath for expressing contiguous elements?

2017-05-01 10:07:18
Here is one without any negation:

/*/A[preceding-sibling::A][following-sibling::A]
           [preceding-sibling::*[1][self::A]]
           [following-sibling::*[1][self::A]]
and true()


On Mon, May 1, 2017 at 7:21 AM, Michael Kay mike(_at_)saxonica(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
not(*[not(self::A)][following-sibling::A][preceding-sibling::A])

returns true if and only if there is no non-A element that is both preceded 
and followed by an A element, which I think is equivalent to your expression 
and rather simpler.

Michael Kay
Saxonica


On 1 May 2017, at 14:00, Costello, Roger L. costello(_at_)mitre(_dot_)org 
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:

Hi Folks,

I want an XPath expression that implements this rule:

      All <A> elements shall be contiguous within <Test>.

In this example all A's are contiguous, so the rule is satisfied:

<Test>
   <B/>
   <A/>
   <A/>
   <A/>
   <B/>
</Test>

In this example there is an intervening B, so the rule is not satisfied:

<Test>
   <B/>
   <A/>
   <A/>
   <B/>
   <A/>
   <B/>
</Test>

Below is the XPath that I created. I have two questions about it:

(1) Is it correct? Do you see anything it would fail to catch?

(2) Is there a better (simpler, more efficient) XPath expression?

Here is the XPath (within a Schematron rule):

<sch:rule context="Test">
   <sch:assert test="
       every $i in A satisfies
          (if ($i/preceding-sibling::A) then 
$i/preceding-sibling::*[1][self::A] else true())
          and
          (if ($i/following-sibling::A) then 
$i/following-sibling::*[1][self::A] else true())
       ">
       All A's shall be contiguous within Test.
   </sch:assert>
</sch:rule>





-- 
Cheers,
Dimitre Novatchev
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