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 __________________________________________________ Do You Yahoo!? Send FREE Valentine eCards with Yahoo! Greetings! http://greetings.yahoo.com XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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