Subject: [xsl] getting the node position in source xml in a variable From: Gurvinder Singh <GurvinderS@xxxxxxxxxx> Date: Wed, 27 Feb 2002 12:22:12 +0200 |
Hi I have a template which takes two node-sets as parameters. I loop thru one of these nodeset using for-each and output the node value along with the corresponding node in the second node-list. Now the problem is that it works fine if i dont have a sort but with sort only the first node list is sorted. So it takes the wrong corresponding values from the second node-list. because the position gives the position in the context. So i should use <xsl:number> but when i put the xsl:number in the variable it always gives 1. even the position() if inside <xsl:variable> gives always 1, but if used in select attribute of <xsl:variable> it works fine <xsl:template name="lookup"> <xsl:param name="name"/> <xsl:param name="datanodes"/> <xsl:param name="displaynodes"/> <select> <xsl:attribute name="name"><xsl:value-of select="$name"/></xsl:attribute> <xsl:for-each select="msxsl:node-set($displaynodes)"> <xsl:sort select="."></xsl:sort> <xsl:variable name="num" select="position()"/> <xsl:variable name="data"><xsl:value-of select="msxsl:node-set($datanodes)[$num]"/></xsl:variable> <option id="{$data}" value="{$data}"> <xsl:value-of select="$data"/>-<xsl:value-of select="."/> </option> </xsl:for-each> </select> </xsl:template> Thanks & Regards Gurvinder Amdocs Limited , Cyprus XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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