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[xsl] Re: How to call a template that can generate a file name from an element

Subject: [xsl] Re: How to call a template that can generate a file name from an element
From: Ian Lang <ianplang@xxxxxxxxx>
Date: Sat, 17 Jan 2004 08:38:35 -0800 (PST)

Thanks for the help so far.  I still do not see what
is causing my problem.  In the full example I
origonally sent the the template looks like this
(actual file name generation commented out until I can
get just the name to work):
  <xsl:template name="createBaseFrameFileName">
    <xsl:param name="element"/>
    <xsl:value-of select="$element[@name]"/>
 <!--    <xsl:value-of select="concat($outputDir,
$fileSep, $element[@name], '_index', '.html')"/> -->

and it gets used to create a variable like this:
    <xsl:variable name="generatedName">
        <xsl:with-param name="element" select="."/>

I like the idea of using a default for the param
element to use the context node - thanks.  But it
still does not want to pass the context element as a
parameter for some reason.  My select looks like your
example and my XPath expression to get the name looks
good.  I must be missing something!



David Carlisle <davidc@xxxxxxxxx> wrote
>  Since passing the element
>  as a parameter and then using '$element[@name]' did
>  not work 
> If you passed an element as the parameter then 
> $element[@name] would
> work. If it didn't work then presumably you did not
> do that, perhaps you
> passed a string with an element name 
> instead? 'foo'[@name] is not valid Xpath.

Wendell Piez <wapiez@xxxxxxxxxxxxxxxx> wrote

At 12:29 PM 1/14/2004, you wrote:
> It will work: we do this all the time. Sometimes 
> it's even recommended to 
> do this even when you don't have to, instead of 
> relying on context 
> implicitly (it's more versatile and clearer for 
> maintenance).
> As I said earlier, check whether $element[@name] 
> really gets you what you 
> want, and not something else. $element[@name] is not

> the same as $element/@name. :->

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