Re: accessing different nodes when comparing two xml

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Subject: Re: accessing different nodes when comparing two xml files
From: "Aron Bock" <aronbock@xxxxxxxxxxx>
Date: Mon, 25 Apr 2005 16:09:24 +0000

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:  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

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Current Thread
accessing different nodes when comparing two xml files RahilQ - 25 Apr 2005 13:31:57 -0000 David Carlisle - 25 Apr 2005 13:48:03 -0000 RahilQ - 25 Apr 2005 14:39:04 -0000 David Carlisle - 25 Apr 2005 15:00:55 -0000 Aron Bock - 25 Apr 2005 16:09:47 -0000 <=

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Re: accessing different nodes, David Carlisle	Thread	First attempt at xsl:result-d, Spencer Tickner
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