Subject: Re: testing for last node in a list From: Ann Marie Rubin - Sun PC Networking Engineering <Annmarie.Rubin@xxxxxxxxxxxx> Date: Tue, 23 May 2000 10:44:20 -0400 (EDT) |
Jeni, Thanks. It worked! Ann Marie X-Sender: JTennison@NTServer To: Annmarie.Rubin@xxxxxxxxxxxx From: Jeni Tennison <Jeni.Tennison@xxxxxxxxxxxxxxxx> Subject: Re: testing for last node in a list Cc: xsl-list@xxxxxxxxxxxxxxxx Mime-Version: 1.0 X-MDaemon-Deliver-To: Annmarie.Rubin@xxxxxxxxxxxx Content-Transfer-Encoding: 8bit X-MIME-Autoconverted: from quoted-printable to 8bit by purol.East.Sun.COM id GAA17422 Ann, >I am generating a list of ancestor nodes for a matched CLASS element. The XSL >calls this template to output the ancestors when a CLASS is matched. I want to >output a "|" character after each CLASS node, EXCEPT the last one. I am unable >to express the correct test for the last node in this list. I tried using ><xsl:if test="position()=last()">, but this statement returns true each time the >template is called. > >Is there another way to solve this? Try thinking in terms of 'if I just know about node X, what is is about node X that affects whether or not there is a | above or below me?' The answer is that all nodes have a | above them apart from the one at the top of the hierarchy. How can you tell if you are at the top of the hierarchy? Because you don't have a parent node. So, try: <!-- named template to do the hierarchy tracing --> <xsl:template match="CLASS" mode="hierarchy"> <xsl:if test="@SUPERCLASS"> <xsl:apply-templates select="key('classes', @SUPERCLASS)" mode="hierarchy"/> <xsl:text>|</xsl:text> </xsl:if> <br data="{@SUPERCLASS}"> <a href="{@NAME}.html"><xsl:value-of select="@NAME"/></a> </br> </xsl:template> Hope that helps, Jeni Dr Jeni Tennison Epistemics Ltd, Strelley Hall, Nottingham, NG8 6PE Telephone 0115 9061301 ? Fax 0115 9061304 ? Email jeni.tennison@xxxxxxxxxxxxxxxx XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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