Re: [xsl] Processing IDREFS attributes

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Hi,

I'd like to know which processor Dan is running.

At 10:52 PM 11/1/2005, Ken wrote:
> > > then you should be able to say:
> > >
> > >   <xsl:for-each select="id(@references)/@id">
> > >     <xsl:if test="position()>1"> </xsl:if>
> > >     <xsl:value-of select="."/>
> > >   </xsl:for-each>
> >
> > A slight twist on this does what I wanted, which was to go find the 
> > element that has the ID:
> >
> > >   <xsl:for-each select="id(@references)/@id">
> > >     <xsl:value-of select="id(.)"/>
> > >   </xsl:for-each>
> >
> > This produces the content of the element with the assigned ID. 
> > Without it, I just got the id value.

This could be

<xsl:for-each select="id(@references)">
  <xsl:value-of select="."/>
</xsl:for-each>

couldn't it? Since you're no longer reporting the value of the 
attribute (as Ken was), no need to traverse there (and then call id() again).

> > With my original code with added context:
> > <xsl:templete match='foo'>
> > <xsl:for-each select="id(@references)">
> >         <xsl:value-of select="."/>
> > </xsl:for-each>
> > </xsl:template>
> >
> > It didn't loop for one problem, so there was only a single output, 
> > instead of two values as I expected. The output was just the same 
> > value as my @references content.
>
> I can't fathom what is going on and why the @references is just 
> being repeated ... that code should have given you the concatenated 
> string values of the elements whose ID typed attribute has a value 
> in the tokenized set of values of the @reference node.

That's why I'd like to know which processor this is, to try and reproduce this.

Cheers,
Wendell



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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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