Subject: Re: [xsl] most efficient way to get XML source's parent dir path From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Tue, 10 Feb 2009 11:55:52 -0500 |
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?
-- Upcoming hands-on XSLT, UBL & code list hands-on training classes: Brussels, BE 2009-03; Prague, CZ 2009-03, http://www.xmlprague.cz Training tools: Comprehensive interactive XSLT/XPath 1.0/2.0 video Video lesson: http://www.youtube.com/watch?v=PrNjJCh7Ppg&fmt=18 Video overview: http://www.youtube.com/watch?v=VTiodiij6gE&fmt=18 G. Ken Holman mailto:gkholman@xxxxxxxxxxxxxxxxxxxx Crane Softwrights Ltd. http://www.CraneSoftwrights.com/s/ Male Cancer Awareness Nov'07 http://www.CraneSoftwrights.com/s/bc Legal business disclaimers: http://www.CraneSoftwrights.com/legal
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
[xsl] most efficient way to get XML, Robert Koberg | Thread | Re: [xsl] most efficient way to get, Robert Koberg |
[xsl] most efficient way to get XML, Robert Koberg | Date | Re: [xsl] most efficient way to get, Robert Koberg |
Month |