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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, $element[@name], '_index', '.html')"/> -->
</xsl:template>
and it gets used to create a variable like this:
<xsl:variable name="generatedName">
<xsl:call-template
name="createBaseFrameFileName">
<xsl:with-param name="element" select="."/>
</xsl:call-template>
</xsl:variable>
I like the idea of using a default for the param
element to use the context node - thanks. But it
still does not want to pass the context element as a
parameter for some reason. My select looks like your
example and my XPath expression to get the name looks
good. I must be missing something!
Thanks,
Ian
-----
David Carlisle <davidc@xxxxxxxxx> wrote
> Since passing the element
> as a parameter and then using '$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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