Re: [xsl] Extract input filename

Subject: Re: [xsl] Extract input filename
From: "J. S. Rawat" <jrawat@xxxxxxxxxxxxxx>
Date: Fri, 18 May 2007 09:32:24 +0530
Thanks Abel and Micheal, thanks a lot!!!. Here mistake was mine. I was not supplying parameter (meta="input_meta.xml") in the command line yesterday. I have just saw your post in my mailbox and tried once again by fresh mind and got the result in the first time. Thanks and thanks a lot.
Thanks and Regards
...JSR


At 04:10 PM 5/17/2007 +0200, Abel Braaksma wrote:
J. S. Rawat wrote:
At 11:10 AM 5/17/2007 +0100, you wrote:
Well, that probably means you changed the parameter name from "meta" to
"data" in one place but not in the other place.

Please have look over following. Is it right? But the error is same.


<xsl:param name="meta" as="xs:string" required="yes"/>
<xsl:variable name="filename" select="document($meta)" />
<xsl:template match="/">
        <doi>
      <xsl:copy-of select="$filename/doi/text()" />
        </doi>
</xsl:template>

that is not possible: you cannot have an error saying that you did not supply a value for parameter 'data' and not have a parameter data in your stylesheet. You are not showing the relevant parts. If you also have a parameter 'data', then you have to set that one too.


For a follow-up, please create minimal example that includes a complete stylesheet and the command line you used. Test the example before you post it and copy and paste the exact error you get.

-- Abel

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