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Re: arguments for xsl:call-template
- From: "Thomas B. Passin" <tpassin at mitretek dot org>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Fri, 7 Dec 2001 10:20:01 -0500
- Subject: Re: [xsl] arguments for xsl:call-template
- References: <5.1.0.14.0.20011207155637.00a03ec0@globalia-sistemas.com>
- Reply-to: xsl-list at lists dot mulberrytech dot com
[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.
>
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