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RE: [xsl] most efficient way to get XML source's parent dir path

2009-02-10 12:13:27
last() is a number >= 1, so boolean(last()) is always true, and not(last())
is always false. So a filter expression using the predicate [not(last())]
will always deliver an empty sequence.

Use [position() ne last()]

Michael Kay
http://www.saxonica.com/ 

-----Original Message-----
From: Robert Koberg [mailto:rob(_at_)koberg(_dot_)com] 
Sent: 10 February 2009 16:53
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] 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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