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RE: [xsl] Extract input filename

2007-05-17 00:47:21

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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