Subject: Re: [xsl] correction: how to get new position() of a sorted result tree From: Samuzeau Pascal <samuzeau@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> Date: Mon, 06 Aug 2001 09:50:31 +0200 |
Hi, I have the same problem, and I work with Cocoon 1.8. Is there any answer ? PS Michael Kay wrote: > > I apologize for the typo in my previous question. The > > original XSLT code > > is: > > <xsl:template match="/"> > > <xsl:apply-templates select="//article"> > > <xsl:sort data-type="number" > > order="descending" select="@date" /> > > </xsl:apply-templates> > > </xsl:template> > > > > <xsl:template match="article"> > > <xsl:if test="position() < 100"> > > ...... <!-- do processing here --> > > </xsl:if> > > </xsl:template> > > > > > > The question is how to get the position() from a sorted > > result tree. In > > the above code, calling position() returns the ID of the > > original pre-sorted > > tree. > > It should return the position in the sorted sequence. > > Which processor are you using? > > Mike Kay > Software AG > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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