Subject: Re: [xsl] getting the node position in source xml in a variable From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Wed, 27 Feb 2002 13:55:28 +0000 |
Hi Gurvinder, I'm afraid that I can't tell what's going wrong without you supplying some more information, specifically about how you're calling the template. I tried using this source XML: <test> <datanodes> <data>3</data> <data>2</data> <data>1</data> </datanodes> <displaynodes> <display>C</display> <display>B</display> <display>A</display> </displaynodes> </test> With this call to the template: <xsl:call-template name="lookup"> <xsl:with-param name="name" select="'test'" /> <xsl:with-param name="datanodes" select="/test/datanodes/data" /> <xsl:with-param name="displaynodes" select="/test/displaynodes/display" /> </xsl:call-template> And I got this result: <select name="test"> <option id="1" value="1">1-A</option> <option id="2" value="2">2-B</option> <option id="3" value="3">3-C</option> </select> One problem that could be occurring is if the $displaynodes aren't actually siblings of each other. But without seeing how you're actually calling the template and what the source of your stylesheet looks like, I can't tell. Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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