good point however use
empty($seq)…
and not count
On 25.03.2015, at 19:52, Wolfgang Laun
wolfgang(_dot_)laun(_at_)gmail(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
On 25 March 2015 at 18:36, Dimitre Novatchev dnovatchev(_at_)gmail(_dot_)com
<mailto:dnovatchev(_at_)gmail(_dot_)com>
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com
<mailto:xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com>> wrote:
every $v in 1 to (count($sequence)-1) satisfies ($sequence[$v] lt
$sequence[$v+1])
This may be inefficient (depending on the XPath engine being used).
I think the following has better chances of being more efficient --
write an xsl:function, name it, say, "increasing". The body of this
function can be just:
not($seq[2]) or $seq[1] lt $seq[2] and increasing(subsequence($seq, 2))
It would appear that this makes -1,0,2,1 an ascending sequence. I dislike
this "empty-is-false trick" - once in a while you get bitten.
count($seq) < 2 or ...
-W
Cheers,
Dimitre
On Wed, Mar 25, 2015 at 10:20 AM, Leo Studer
leo(_dot_)studer(_at_)varioweb(_dot_)ch
<mailto:leo(_dot_)studer(_at_)varioweb(_dot_)ch>
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com
<mailto:xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com>> wrote:
A similar problem as before, is the integer sequence increasing?
this is my solution:
every $v in 1 to (count($sequence)-1) satisfies ($sequence[$v] lt
$sequence[$v+1])
Do you have a better one?
Cheers
Leo
--
Cheers,
Dimitre Novatchev
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