[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
Thanks Ken, Andrew and Mike .
I don't know why I had blinders on regarding the replace function...
On Feb 10, 2009, at 11:55 AM, G. Ken Holman wrote:
Re: [xsl] most efficient way to get XML source's parent dir path
Subject: Re: [xsl] most efficient way to get XML source's parent dir path From: Robert Koberg <rob@xxxxxxxxxx> Date: Tue, 10 Feb 2009 11:59:04 -0500 |
Thanks Ken, Andrew and Mike .
I don't know why I had blinders on regarding the replace function...
thanks, -Rob
On Feb 10, 2009, at 11:55 AM, G. Ken Holman wrote:
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@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 -> |
---|---|---|
Re: [xsl] most efficient way to get, G. Ken Holman | Thread | RE: [xsl] most efficient way to get, Michael Kay |
Re: [xsl] most efficient way to get, G. Ken Holman | Date | RE: [xsl] most efficient way to get, Michael Kay |
Month |