At 2009-02-10 11:53 -0500, Robert Koberg wrote:
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?
How about replacing the last component of the URI with nothing:
select="replace(document-uri(/),'/[^/]*$','')"
I hope this helps.
. . . . . . . . . . Ken
--
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(_at_)CraneSoftwrights(_dot_)com
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
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--