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Re: [xsl] arguments for xsl:call-template
Subject: Re: [xsl] arguments for xsl:call-template From: "Thomas B. Passin" <tpassin@xxxxxxxxxxxx> Date: Fri, 7 Dec 2001 10:20:01 -0500 |
[Pep Coll] > I don't understand quite well what you want, because you can do this: > <xsl:with-param name="path" select="'/report/histo/bar'" /> > but you can do this , > <xsl:with-param name="path" select="'/report/histo/@bar'" /> NO, @bar would only return attributes named "bar" and the original example had "bar" elements, not attributes. Tom P > because you are not assigning anything to var. path and also you are doing > histogram just once so the 'for-each' has no reason to be. Explain what's > the purpose of this template. > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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