Re: [xsl] Increasing sequence ?

["Dimitre Novatchev dnovatchev@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

> 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))


Cheers,
Dimitre


On Wed, Mar 25, 2015 at 10:20 AM, Leo Studer leo.studer@xxxxxxxxxxx
<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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