[XSL-LIST Mailing List Archive Home] [By Thread] [By Date]

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


Subject: Re: [xsl] Using Position() to display results in groups
From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx>
Date: Mon, 20 Aug 2001 15:43:53 +0100

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/


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list



Current Thread
Keywords
xml