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Re: [xsl] Divide to pages


Subject: Re: [xsl] Divide to pages
From: "andrew welch" <andrew.j.welch@xxxxxxxxx>
Date: Wed, 5 Apr 2006 16:12:49 +0100

On 4/5/06, Paull <paullus4mlist@xxxxxxxxx> wrote:
> Hello All,
> following xml:
> <data>
> <item name="1" id="i1">v1</item>
> <item name="2" id="i1">v2</item>
> <item name="3" id="i1">v1</item>
> <item name="4" id="i1">v1</item>
> <item name="5" id="i1">v2</item>
> <item name="6" id="i1">v1</item>
> <item name="7" id="i1">v1</item>
> <item name="8" id="i1">v2</item>
> <item name="9" id="i1">v1</item>
>
> <item name="10" id="i2">v2</item>
> <item name="11" id="i2">v2</item>
> <item name="12" id="i2">v2</item>
>
>
> <group name="g1" id="i1"/>
> <group name="g2" id="i2"/>
> </data>
> should be transformed to the xml, where items are grouped by id, and
> divided to pages whith 5 items per page. Result should be like following:
> g1
> 1. i1v1
> 2. i1v2
> 3. i1v1
> 4. i1v1
> 5. i1v2
> EOP
> 1. i1v1
> 2. i1v1
> 3. i1v2
> 4. i1v1
> g2
> 5. i2v2
> EOP
> 1. i2v2
> 2. i2v2
>
> I can group it, but how to divide for pages - no idea ...

It's difficult to give an exact answer without seeing how you are
grouping, but this can be achieved using mod, eg position() mod 5 = 0
will give you every fifth item.


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