Rahil, as a prior respondent alluded to, document('foo.xml') returns the
root node, "/". See the following diagram:
/ (root node)
|
+--Top
|
+--SubConcepts
| |
| +-- SubConcept
|
+--SubConcepts
As you can see, the root node, "/" does not have any siblings.
By the way, if you're looking for tree comparision routines, you may want to
see Sal Mangano's "XSLT Cookbook" from O'Reilly. It's XSL templates are
available for download on the web, and can be used for constrained
comparisions. Search this NG's archives.
Regards,
--A
From: RahilQ <qamar_rahil(_at_)yahoo(_dot_)co(_dot_)uk>
Reply-To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] accessing different nodes when comparing two xml files
Date: Mon, 25 Apr 2005 15:37:45 +0100
I corrected the <xsl:for-each> condition such that its now
<xsl:for-each
select="document($first)/following-sibling::Top/SubConcepts/SubConcept[(_at_)name=$firstName]/Value>
But it still returns null.
For testing purpose I changed the condition to
<xsl:value-of select="document($arch)/Top/SubConcepts/SubConcept[1]">
and it returned the right value.
I have a feeling its got something to do with the way Ive written the
following-sibling statement.
Any suggestions?
Thanks
Rahil
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