Re: [xsl] getting associated file name for element

[Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>]

Hi Matthew,

>   If I am doing something like:
>
>     <xsl:for-each select="document($all-questions//quiz/@href)//question">
>         <xsl:sort order="ascending" select="faa-num"/>
>       ...do stuff with each question...
>          <xsl:value-of select="faa-num"/>
>          Here's where I want to grab the name of the file, i.e. @href, the
> question came from
>     </xsl:for-each>
>
> where quiz/@href contains the filenames containing questions to be
> grabbed and then sorted by a question child element called faa-num.
> How can I grab the value of @href as each question is evaluated?

Since you need to sort the questions, you can't. I think that the best
way to do what you're trying to do here is to create an intermediate
node tree and use a node-set() extension function to get at the nodes.
So something like:

  <xsl:variable name="questions-rtf">
    <xsl:for-each select="$all-questions//quiz/@href">
      <quiz href="{.}">
        <xsl:copy-of select="document(.)//question" />
      </quiz>
    </xsl:for-each>
  </xsl:variable>
  <xsl:for-each select="exsl:node-set($questions-rtf)/quiz/question">
    <xsl:sort order="ascending" select="faa-num" />
    ... do stuff with each question ...
    <xsl:value-of select="faa-num" />
    ...
    <xsl:value-of select="../@href" />
  </xsl:for-each>

Cheers,

Jeni

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


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