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Re: value of xsl:param in xsl:sort
- From: alex <shortestpath at yahoo dot com>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Wed, 13 Feb 2002 21:16:27 -0800 (PST)
- Subject: Re: [xsl] value of xsl:param in xsl:sort
- Reply-to: xsl-list at lists dot mulberrytech dot com
I had this problem a few days ago:
*[name() = $sortByElement]
--- "Kunal H. Parikh" <kunal@kaypee.net> wrote:
> Hi !
>
> ========================
> <xsl:param name="sortByElement">AuthorList/Author/Name</xsl:param>
> <xsl:sort select="$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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