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() < $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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From: Martin Honnen <Martin(_dot_)Honnen(_at_)gmx(_dot_)de>
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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() < $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/
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