Re: How to call a template that can generate a file na

Oops don't I feel like a heel.  After reading more of
the list and looking at some other samples I realized
I was making it too hard.  I just switched to:
  <xsl:template name="createBaseFrameFileName">
    <xsl:value-of select="concat($outputDir,
        $fileSep, @name, '_index', '.html')"/>
  </xsl:template>
 
And calling it thus:
  <xsl:variable name="generatedName">
    <xsl:call-template 
          name="createBaseFrameFileName"/>
  </xsl:variable>

At the moment it does not look like I will have to
pass another element to this method (other than the
context) but what if I do?  Since passing the element
as a parameter and then using '$element[@name]' did
not work is there another way?

Thanks,

Ian

--- Ian Lang <ianplang@xxxxxxxxx> wrote:
<snip>
>   <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>
> 
>   <!-- Generate the top level html file which sets
> up
> the frames -->
>   <xsl:template match="cabinet|content"
> mode="CreateBaseFrame">
>     <xsl:param name="title"/>
>     <xsl:param name="generator" select="'Our
> Generator'"/>
>     <xsl:param name="description" select="''"/>
>     <xsl:param name="keywords" select="''"/>
> 
>     <xsl:variable name="generatedName">
>       <xsl:call-template
> name="createBaseFrameFileName">
>         <xsl:with-param name="element"
select="."/>
>       </xsl:call-template>
>     </xsl:variable>
<snip>

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