Re: [xsl] Using Position() to display results in groups

[Jeni Tennison <mail@xxxxxxxxxxxxxxxx>]

Hi Andrew,

> Writing ten or so 'when' tests probably isnt the best way to go
> about this. What is required is some code to iterate through the
> list, decide if it as the 10th item ( or a multiple of 10) and
> output some code.

I think that you want to group a load of items into card elements by
position. I would select the first of each group of 10 to process (the
1st, 11th, 21st etc.), using the mod operator to select them:

  <xsl:for-each select="item[position() mod 10 = 1]">
    <card id="{position()}">
      ...
      <a href="#{position() + 10}">next 10</a>
    </card>
  </xsl:for-each>

Within that xsl:for-each, to get the 10 items that should be displayed
on the card, you want this item and its following 9 siblings, e.g.:

  <xsl:for-each select=".|following-sibling::item
                            [position() &lt; 10]">
    ...
  </xsl:for-each>

Actually, I'd be a bit wary of using numbers for IDs like that -- IDs
in XML should start with a letter, and you might have validity
and linking problems if you use something a number. You can just do:

  <card id="card{position()}">
    ...
  </card>

instead.

Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


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