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="//*[(_at_)id='id2']" />
And the following to be true:
generate-id(document('#id2')) = generate-id(//*[(_at_)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(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Richard Lander
Sent: October 8, 2002 2:00 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] 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(_at_)adamvandenhoven(_dot_)com]
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
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