[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] most efficient way to get XML source's parent dir path
Hi,
I want to get the path to the parent directory of the XML used as the source in a transformation. Is this the best way? <xsl:variable name="path-tokens" select="tokenize(document-uri(/), '/')" as="xs:string*"/> <xsl:variable name="source-dir-path" select="string-join(remove($path-tokens, count($path-tokens)), '/')" as="xs:string"/> I also unsuccessfully tried to use only one variable: <xsl:variable name="source-dir-path-test" select="string-join(tokenize(document-uri(/), '/')[not(last())], '/')" as="xs:string"/> which results in an empty string. Is there a better way? thanks, -Rob
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