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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
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