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Re: [xsl] Help calling templates with parameters
Subject: Re: [xsl] Help calling templates with parameters From: "cking" <anton@xxxxxxxx> Date: Wed, 22 Sep 2004 09:32:53 +0200 |
Charlie Consumer wrote: > > and once I enter the template the parameter isn't a > node set like it was before I called the template, but > a node fragment. I don't know what a node fragment > is. Can anyone explain this? A result tree fragment (RTF) is what you get if you fill a param or variable inside the element, instead of in its select attribute. Like you did in call-template: > <xsl:with-param name="currentRule"> > <xsl:value-of select="current()"/> > </xsl:with-param> In XSLT 1.0, you can't use an RTF in XPath expressions. You would have to use an extension, eg. from from EXSLT (www.exslt.org): <xsl:value-of select="exsl:node-set($currentRule)/blah/@id"/> <!-- you can't: <xsl:value-of select="$currentRule/blah/@id"/> --> HTH, Anton
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