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Re: [xsl] Increasing sequence ?

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good point however use

 empty($seq)

and not count


> On 25.03.2015, at 19:52, Wolfgang Laun wolfgang.laun@xxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> On 25 March 2015 at 18:36, Dimitre Novatchev dnovatchev@xxxxxxxxx
<mailto:dnovatchev@xxxxxxxxx> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx
<mailto:xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>> 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) &lt; 2 or ...
>
> -W
>
>
>
>
> Cheers,
> Dimitre
>
>
> On Wed, Mar 25, 2015 at 10:20 AM, Leo Studer leo.studer@xxxxxxxxxxx
<mailto:leo.studer@xxxxxxxxxxx>
> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx
<mailto:xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>> 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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