Subject: Re: [xsl] getting the node position in source xml in a variable From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Wed, 27 Feb 2002 14:20:08 +0000 |
Hi Gurvinder, > That excatly is the problem... > all the nodes in the $displaynodes are not siblings... Ahh... > coz i have something like this > <test> > <nodes> > <level2> > <data>3</data> > <display>C</display> > <someotherdata>abc</someotherdata> > </level2> > <level2> > <data>2</data> > <display>B</display> > <someotherdata>abc</someotherdata> > </level2> > <level2> > <data>1</data> > <display>A</display> > <someotherdata>abc</someotherdata> > </level2> > </nodes> > </test> If the data and display elements that you want to associate are already associated (as they are in this structure) then why not just use that association? For example, rather than using the lookup template at all, you could just do: <select name="test"> <xsl:for-each select="/test/nodes/level2"> <xsl:sort select="display" /> <option id="{data}" value="{data}"> <xsl:value-of select="concat(data, '-', display)" /> </option> </xsl:for-each> </select> I guess that you're using the lookup template so that you can create other select elements in the same kind of way elsewhere. Making the lookup template absolutely generic will make it comparatively inefficient, so before going to that extreme, perhaps you could describe the variety of situations in which you want to use the lookup template? For example, are the display nodes and associated data nodes always siblings of each other, nested under some other node? Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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