Re: [xsl] Getting position of parent

Subject: Re: [xsl] Getting position of parent
From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>
Date: Thu, 20 Feb 2003 15:36:39 +0000
Hi Geoff,

> So what I want to do is make $pageNumber = position() -1 for the
> page node while I am in the (grandchild) Field node. How can I
> accomplish this? I'm scratching my head on this one

Try counting the number of preceding sibling Page elements the
ancestor Page element has and adding one to get the position:

  count(ancestor::Page/preceding-sibling::Page) + 1

By the way, you might find your template a bit clearer if you used an
attribute value template to create the style attribute:

<xsl:template match="Field">
  <xsl:variable name="pageNumber"
    select="count(ancestor::Page/preceding-sibling::Page) + 1" />
  <xsl:variable name="top"
    select="@Top * 0.06666 + ($pageNumber * $pageHeight)" />
  <div style="position: absolute;
              top: {$top};
              left: {@Left * 0.06666};
              font-size: {@FontSize}pt;
              font-family: {@FontFamily};">
    <xsl:value-of select="." />



Jeni Tennison

 XSL-List info and archive:

Current Thread