Hi Everybody,
I tried lot and finally get the solution for the same.
Which give the exact name of file.
Also it full fills the my condition
1] Level of folder may vary. [File should be in any folder on drive.]
please see modified code below...
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="/">
<xsl:text><![CDATA[
]]>
</xsl:text>
<test><xsl:text><![CDATA[
]]>
</xsl:text>
<xsl:for-each select="concept">
<xsl:variable name="filename1"><xsl:value-of select="base-uri()" /></xsl:variable>
<p><xsl:call-template name="name">
<xsl:with-param name="id"><xsl:value-of select="$filename1" /></xsl:with-param>
</xsl:call-template>
</p>
</xsl:for-each>
<xsl:text><![CDATA[
]]></xsl:text>
</test>
</xsl:template>
<xsl:template name="name">
<xsl:param name="id" />
<xsl:choose>
<xsl:when test="$id[contains(.,'\')]">
<xsl:call-template name="name"><xsl:with-param name="id"><xsl:value-of select="substring-after($id, '\')" /></xsl:with-param></xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="substring-before($id, '.')" />
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
If anyone has another idea on this please suggest.
Thanks.
Cheers,
Mahesh
Marked Solved 
Offline 
