RE: Finding the path from the filename

Hi Thomas.
Try this:
 <xsl:template name="FullPath">
  <xsl:param name="PathString" select="''"/>
  <xsl:if test="contains($PathString,'\')">
  <xsl:value-of select="concat(substring-before($PathString,'\'),'\')"/>
  <xsl:call-template name="FullPath">
   <xsl:with-param name="PathString"
select="substring-after($PathString,'\')"/>
  </xsl:call-template>
  </xsl:if>
 </xsl:template>

To call it just do:
  <xsl:call-template name="FullPath">
   <xsl:with-param name="PathString" select="$local_path"/>
  </xsl:call-template>
Where $local_path is a variable or a param with your path string


-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of 
Thomas V.
Nielsen
Sent: segunda-feira, 20 de Janeiro de 2003 15:18
To: XSL List
Subject: [xsl] Finding the path from the filename


In a XSLT I receive a parameter with a full path and file name, like;

C:\Data\Test\File.xml

I have tried fumbling with substring and substring-before, but with no
luck.

What I need is the full path without the file name, like C:\Data\Test\

Sometimes parameter looks like this

..\Test\File.xml

And also here I need to find the path, like ..\Test\

Any suggestions in how to use the substring with some iteration?

/Thomas

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 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list