Dear all,
I have the following two params
<xsl:param name = bfileHrefb select = b
bP:/developers/perf/bigPayload.xmlb
b/>
<xsl:param name="outDir" select="'outDir'"/>

  I want OutFileName to be :
bP:/developers/perf/outDir/bigPayload-Formatted.xmlb

Is there a better way to do this than what I did?
<xsl:variable name="OutFileName" select=" substring-before($fileHref,
tokenize($fileHref,'/')[last() -1 ] ) || $outDir || '/' ||
substring-before( tokenize($fileHref,'/')[last()] ,'.xml') ||
'-Formatted.xml'"/>

thanks.
Dt