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most efficient way to get XML source's parent dir path

Subject: most efficient way to get XML source's parent dir path
From: Robert Koberg <rob@xxxxxxxxxx>
Date: Tue, 10 Feb 2009 11:53:24 -0500
 most efficient way to get XML source's parent dir path
Hi,

I want to get the path to the parent directory of the XML used as the source in a transformation.

Is this the best way?

<xsl:variable
name="path-tokens"
select="tokenize(document-uri(/), '/')" as="xs:string*"/>
<xsl:variable
name="source-dir-path"
select="string-join(remove($path-tokens, count($path-tokens)), '/')" as="xs:string"/>


I also unsuccessfully tried to use only one variable:

<xsl:variable
name="source-dir-path-test"
select="string-join(tokenize(document-uri(/), '/')[not(last())], '/')" as="xs:string"/>


which results in an empty string.

Is there a better way?

thanks,
-Rob

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