Subject: RE: [xsl] Extract input filename From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Thu, 17 May 2007 08:46:48 +0100 |
1. Change this: <xsl:variable name="filename" select="document('input_meta.xml')" /> to this: <xsl:param name="meta" as="xs:string" required="yes"/> <xsl:variable name="filename" select="document($meta)" /> (you might also like to change the name of the variable filename, as it's a misleading name) 2. Change your command line to java -jar saxon8.jar input.xml x.xsl meta=input_meta.xml >c.xml Michael Kay http://www.saxonica.com/ > I have no idea about to extract input filename by passing > argument in command line. In the below examples, I don't want > to hardcode "input_meta.xml". > > Input files > 1. input.xml > 2. input_meta.xml > > command line > java -jar saxon8.jar input.xml x.xsl >c.xml > > Stylesheet > <xsl:variable name="filename" select="document('input_meta.xml')" /> > <xsl:template match="/"> > <doi> > <xsl:copy-of select="$filename/doi/text()" /> > </doi> > </xsl:template> > > Thanks in advance. > JSR
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