XPath 1.0:
not(B[1]/following-sibling::node()[not(position() >
../count(B))][not(self::B)])
XPath 2.0:
not(node()[. >> ../B[1] and . << ../B[last()]][not(self::B)])
Cheers,
Dimitre
On Tue, Jul 19, 2016 at 9:16 AM, G. Ken Holman
g(_dot_)ken(_dot_)holman(_at_)gmail(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
I like Mike's answer for XSLT 2.
Your XSLT 1 solution can be improved by reducing the number of counting
actions. Your solution has three uses of count() ... this has only two:
count(B[last()]/preceding-sibling::*[not(self::B)]) =
count(B[1]/preceding-sibling::*[not(self::B)])
Avoiding the use of counting you could do the following:
string( generate-id( B[1]/preceding-sibling::*[not(self::B)][1] ) ) =
string( generate-id( B[last()]/preceding-sibling::*[not(self::B)][1] ) )
The use of string() is necessary in case the sequence of children starts
with B.
Come to think of it, neither your answer, Mike's answer nor my answer will
work if there are no B children at all. You will get a false positive.
Does your test need to accommodate that situation?
. . . . . . . . Ken
At 2016-07-19 15:44 +0000, Costello, Roger L. costello(_at_)mitre(_dot_)org
wrote:
Hi Folks,
This XML has a solid block of <B> elements:
<Document>
<A/>
<B/>
<B/>
</Document>
This XML has an intervening <C> element:
<Document>
<A/>
<B/>
<C/>
<B/>
</Document>
I need an XPath expression to return a Boolean value:
Return true if there is one solid block of <B> elements
(no intervening elements).
Otherwise, return false.
I created a horrendous XPath expression to solve it:
count(B) eq (B[last()]/count(preceding-sibling::*)+1 -
B[1]/count(preceding-sibling::*))
Can you provide a better (simpler, more efficient) XPath expression?
/Roger
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Cheers,
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