saxon passing directory as a command line parameter to XSLT 2.0 style sheet

Question

Using Saxon9he, I would like to pass the directory of the input file to my XSLT style sheet. I have defined the parameter in my style sheet:

<xsl:param name="publishing.folder" />

and on the command line:

+publishing.folder="%~dp1\"

With the + the parameter is empty.

Without the +,

publishing.folder="%~dp1\" 

I get: Error in xsl:result-document/@href Resolved URL is malformed: unknown protocol: d

The d could be the drive letter, that's where the input file is.

Here is the code with the @href causing the error:

    <xsl:template match="/">
      <xsl:call-template name="write-dataset-file">
            <xsl:with-param name="filename" select="concat($publishing.folder,'-dataset.xml')"/>
      </xsl:call-template>
    </xsl:template>

   <xsl:template name="write-dataset-file">
      <xsl:param name="filename"/>

      <xsl:result-document href="{$filename}" omit-xml-declaration="false" method="xml" indent="yes">
         <Dataset>

         </Dataset>
      </xsl:result-document>
  </xsl:template>

xalan accepts the command-line parameter -PARAM publishing.folder "%~dp1\", but I'd rather not go back to XSLT 1.0

Answer 1

The href attribute of xsl:result-document must be a URI, but you are constructing a filename. Although many XML-related software packages accept filenames where the standards require a URI, Saxon tends to be more strict. The simplest way to fix this is to add "file:///" at the start of the filename, though this might not be adequate if the filename contains special characters such as "#".