I'm using the XLink model so my hrefs are URIs. In the example I gave,
it's a fragment identifier which should be expanded relative to the
source document (I understand xpointer to allow this). I'm assuming that
'#someval' is still a shortcut which points to some element with an id
value of 'someval'. 

>From http://www.w3.org/TR/xslt#function-document:
If the URI reference does not contain a fragment identifier, then a
node-set containing just the root node of the document is returned. If
the URI reference does contain a fragment identifier, the function
returns a node-set containing the nodes in the tree identified by the
fragment identifier of the URI reference.

This suggests to me that if I use <xsl:apply-templates
select="document('#id2')" />, I should get the node of the source
document which has an id attribute equal to 'id2'. In the same way, it
suggests that if I use <xsl:apply-templates
select="document('somefile.xml#id2')" />, I should get the node whose id
is 'id2' in 'somefile.xml' which is a sibling to the source document. 

So my question is assuming that there is a node with an id (in this case
an id attribute) with a value of 'id2' in my source document can I
reasonably expect the following to be equivalent:

<xsl:apply-templates select="document('#id2')" />

<xsl:apply-templates select="//*[@id='id2']" />

And the following to be true:

generate-id(document('#id2')) = generate-id(//*[@id='id2'])

Before I go and start telling excelon that they seem to have a bug in
their product, I want to make sure I'm right in my assumption (if not,
I'm going to have to re-evaluate how I do things). 

Adam

> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx 
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of 
> Richard Lander
> Sent: October 8, 2002 2:00 PM
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: RE:  Using document()
> 
> 
> The XSLT processor is looking for a resolvable path as the 
> argument to the document function.
> 
> If the value of the xlink:href attribute is a path to an 
> actual XML file, then it should work. If it isn't, and it's 
> an IDREF-type link to attribute that contains that value, 
> then you'll need an XPath that will get that value for you.
> 
> w/o more information on your document structure ... It is a 
> little hard on this end ...
> 
> Rich
>  
> -----Original Message-----
> From: Adam van den Hoven [mailto:list@xxxxxxxxxxxxxxxxxxx] 
> Sent: Tuesday, October 08, 2002 1:50 PM
> To: XSL Mailing list
> 
> Sorry if this appears twice, I'm having some problems with my 
> mail systems. 
> 
> I have a question. 
>  
> I want to use the following:
>  
> <xsl:apply-templates select="document('#id2')" /> 
>  
> Ok, I'd never actually use this. In reality its something like:
>  
> <xsl:apply-templates select="document(@xlink:href)" />
>  
> which is likely to contain simply fragment identifiers since 
> I'm using XLinks to normalize a sitemap hierarchy. 
>  
> When I try to run this as part of a transform, I get an error 
> saying that the file (fully resolved to refer to the source 
> doc) cannot be found. Is this a problem with how I'm doing 
> things or perhaps a problem with app I'm using to develop 
> templates with (Stylus Studio)?
>  
> Do you have any other ideas about how I can accomplish the 
> same thing given that some of my xlink:href attributes will 
> point to external documents?
>  
> Thanks, 
> H. Adam van den Hoven
> 
> Web Developer
> 
> Credit Union Central of BC
> 
> 
> __________________________________________________
> D O T E A S Y - "Join the web hosting revolution!"
>              http://www.doteasy.com
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 
> 
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 
> 


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