Subject: RE: [xsl] correction: how to get new position() of a sorted result tree From: "Michael Kay" <mhkay@xxxxxxxxxxxx> Date: Thu, 2 Aug 2001 16:25:16 +0100 |
> 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
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