Re: [xsl] question on EXSLT data partitioning

[Erwan AMOUREUX <erwan.amoureux@xxxxxxxxxxxxxxxxxxxxxxx>]

Fisrt sorry for my english, litlle out of practice

Second, for this data volume , it would be nice to use key, isn't it ?
You cannot use a variable for modulo but that accelerate processing

<xsl:key name="part" match="Entity" use="(position() - (position() mod 1000))
div 1000 ">
[...]
<xsl:for-each select="Entity[position()  mod 1000 = 0]">
      <arguments>
        <xsl:apply-templates select ="key('part',position() div 1000) ">
</arguments>
    </xsl:for-each>

[...]

I can't test (and make always error with modulo and other...) but that's the
scheme

Erwan Amoureux
 ----- Message d'origine -----
De : "Hermann Stamm-Wilbrandt" <STAMMW@xxxxxxxxxx>
@ : <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
Envoyi : mercredi 20 octobre 2010 14:56
Objet : Re: [xsl] question on EXSLT data partitioning


>> It should be possible without recursion
>>    <xsl:for-each select="Entity[(position() - 1) mod $N = 0]">
>>      <arguments>
>>        <xsl:apply-templates select=". |
>> following-sibling::Entity[position() &lt; $n"/>
>>      </arguments>
>>    </xsl:for-each>
>
> Thanks Martin, that is perfect
>
>> But I am not sure whether that is the approach you don't want.
>
> Yes, that is what I wanted.
>
> Mit besten Gruessen / Best wishes,
>
> Hermann Stamm-Wilbrandt
> Developer, XML Compiler, L3
> Fixpack team lead
> WebSphere DataPower SOA Appliances
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> IBM Deutschland Research & Development GmbH
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>
>
>
> From:       Martin Honnen <Martin.Honnen@xxxxxx>
> To:         xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Date:       10/19/2010 08:09 PM
> Subject:    Re: [xsl] question on EXSLT data partitioning
>
>
>
> Hermann Stamm-Wilbrandt wrote:
>
>> yesterday I was asked by a colleague on data partitioning.
>> He wanted to partition 100000s of Entities in blocks of 1000
>> for sending a single Database update for 1000 entities.
>>
>> Below is the simplified input, partition size is N=3 and the
>> requested output. Below that is the solution I provided.
>>
>> Here are my questions:
>> * can this task be done without recursion in EXSLT?
>>   [the colleage did not like the idea of doing the partitioning with
>>    just XPath (1<=position()<=1000, 1001<=position()<=2000, ...)
>>    because of the 6 digit number of entities]
>
> It should be possible without recursion
>   <xsl:for-each select="Entity[(position() - 1) mod $N = 0]">
>     <arguments>
>       <xsl:apply-templates select=". |
> following-sibling::Entity[position() &lt; $n"/>
>     </arguments>
>   </xsl:for-each>
> But I am not sure whether that is the approach you don't want.
>
>
> --
>
>             Martin Honnen
>             http://msmvps.com/blogs/martin_honnen/