Re: [xsl] Divide to pages

[Paull <paullus4mlist@xxxxxxxxx>]

andrew welch wrote:

> 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.
>
>  
I'm grouping with following:

<xsl:template match="/data">
  <xsl:for-each select="group">
     <xsl:variable name="id" select="@id"/>
     <b><xsl:value-of select="$id"/></b><br/>
      <xsl:apply-templates select="/data/item[@id=$id]" />
  </xsl:for-each>   
</xsl:template>

<xsl:template match="item">
  <xsl:value-of select="@id"/> <xsl:value-of select="."/><br/>
<xsl:if test="position() mod 5 = 0">
   <!-- EOP -->
</xsl:if>
</xsl:template>

But using <xsl:if test="position() mod 5 = 0"> is not correct because it 
not handles situation with 5 groups each with one item should be pasted 
into one page.