Re: [xsl] accessing different nodes when comparing two xml files

["Aron Bock" <aronbock@xxxxxxxxxxx>]

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@xxxxxxxxxxx>
Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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($first)/following-sibling::Top/SubConcepts/SubConcept[@name=$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

_________________________________________________________________
Is your PC infected? Get a FREE online computer virus scan from McAfee. 
Security. http://clinic.mcafee.com/clinic/ibuy/campaign.asp?cid=3963