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Re: [xsl] value of xsl:param in xsl:sort

Subject: Re: [xsl] value of xsl:param in xsl:sort
From: alex <shortestpath@xxxxxxxxx>
Date: Wed, 13 Feb 2002 21:16:27 -0800 (PST)
I had this problem a few days ago:
*[name() = $sortByElement]

--- "Kunal H. Parikh" <kunal@xxxxxxxxxx> wrote:
> Hi !
> 
> ========================
> <xsl:param name="sortByElement">AuthorList/Author/Name</xsl:param>
> <xsl:sort select="$sortByElement" order="ascending"></xsl:sort>
> ========================
> 
> The above code does not seem to be replacing the $sortByElement for the
> parameter.
> 
> But, if I replace "$sortByElement" with "AuthorList/Author/Name",
> everything
> works out just fine.
> 
> Can someone please suggest what mistake I am making ?
> 
> 
> 
> TIA,
> 
> Kunal

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