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Re: [xsl] Increasing sequence ?


Subject: Re: [xsl] Increasing sequence ?
From: "Leo Studer leo.studer@xxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Thu, 26 Mar 2015 18:26:58 -0000

Dimitre, yes, this is nice, however can not be used in xsd:assertions.
And there is a stack problem with large sequences.

Thanks anyway
Leo

> On 25.03.2015, at 18:36, Dimitre Novatchev dnovatchev@xxxxxxxxx
<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))
>
>
> 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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