Re: [xsl] remove relative path

Subject: Re: [xsl] remove relative path
From: "Jon Gorman" <jonathan.gorman@xxxxxxxxx>
Date: Tue, 16 May 2006 20:30:07 -0400
On 5/16/06, Gav.... <brightoncomputers@xxxxxxxxxxxxxxxxxxx> wrote:
Hi All,

I have looked through the list, archives etc and come up some code based on
what I summized from it, but of course it does not work.

For an OS project, I'm trying to remove all ../ from the supplied @src
Obviously they can be nested values too such as ../../../

So I need to remove all these leaving just the filename. (I can then
A predetermined path onto it.)

I have a template :-

<xsl:template name="removedotdots">
          <xsl:param name="path"/>
          <xsl:variable name="removedirs"
          <xsl:variable name="removeagain"
          <xsl:if test="$removeagain">
                  <xsl:call-template name="removedotdots">
                          <xsl:with-param name="path"

But what if it's five levels? Fifteen? Why do this approach? Recursion is your friend with XSLT.

XSLT 2.0 has better ways, and probably so does XSLT 1.0.  But right
off the top of my head even without thinking, I can come up with a
recursive way:

Something like:

<xsl:template match="path">

<xsl:call-template name="getFilename">
<xsl:with-param name="path_string" select="." />


<xsl:template name="getFilename">
<xsl:param name="path_string" />
<xsl:when test="starts-with($path_string,'../')">
	<xsl:call-template name="getFilename">
	<xsl:with-param	name="path_string"
select="substring-after($path_string,'../')" />
<xsl:when test="starts-with($path_string,'/../')">
<xsl:call-template name="getFilename">
<xsl:with-param name="path_string"
select="substring-after($path_string,'/../')" />
<xsl:value-of select="$path_string"/>


This assuming the path is an element in path, but it should be
straightforward enough to figure out.

There's probably lots of other ways to do this.

Jon Gorman

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