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Re: [xsl] Processing IDREFS attributes

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below

At 02:23 PM 11/1/2005, you wrote:

*If* you have properly declared the id attributes, for my example they are called id=, something along the lines of:

<!ATTLIST foo id ID #IMPLIED>

Yes and I had the references defined as IDREFS #IMPLIED as well.

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. I'm not sure why you have the position test, can you explain that?

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.

In my original code I had this as well, trying to get to the content of the element referenced, but nothing was being output, so I used the current value just to see if anything was being looked up.

<xsl:value-of select="id(.)"/>

and it would report a space-separated list of unique token values (rather than just @references since that might have duplicates).

If you don't have a declaration of the attribute type, then id() has nothing to look up and will always return an empty node set.

Yep, had that problem covered.

I hope this helps, Dan.

Yes, once again you have come to the rescue. I sort of see why adding the /@id to the for-each select worked, but I don't know why mine didn't work or like you said didn't give me all the content.

..dan
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Danny Vint

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