Re: telling an outer template which template to call..

See:

  "The Functional Programming Language XSLT - A proof through examples"
at


http://fxsl.sourceforge.net/articles/FuncProg/Functional%20Programming.html

or

"Functional programming in XSLT using the
FXSL library"

at


http://www.mulberrytech.com/Extreme/Proceedings/xslfo-pdf/2003/Novatchev01/EML2003Novatchev01.pdf



=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL



"Felix Breuer" <felix@xxxxxxxxxx> wrote in message
news:1062447181.9215.24.camel@xxxxxxxx
> Hello!
>
> I would like an outer template to call an inner template which is
> specified by a parameter. I.e. I would like to get the following example
> working:
>
> <xsl:template match="/">
>   <xsl:apply-templates select="foo">
>     <xsl:with-param name="para">bar</xsl:with-param>
>   </xsl:apply-templates select="foo">
> </xsl:template>
>
> <xsl:template match="foo">
>   <xsl:param name="para">moo</xsl:with-param>
>   ...
>   <xsl:call-template name="{$para}"/>
>   ...
> </xsl:template>
>
> <xsl:template name="bar">...</xsl:template>
> <xsl:template name="moo">...</xsl:template>
>
> I hope the above makes it clear what I am aiming at. [Obviously the
> whole stylesheet is going to be a bit more complicated or such an
> involved solution wouldn't be necessary.]
>
> Now, when I run the above through xsltproc (libxslt) I get errors like
>
> xsl:call-template : template {$para} not found
>
> Obviously, the problem is that the parameter $para is not expanded
> before the call-template element is executed.
>
> My questions are:
> 1) Is $para supposed not to be expanded in time? Or is this a bug in
> libxslt?
> 2) Is there a way to get this stylesheet working? With libxslt or any
> other processor?
>
> Thanks in advance,
> Felix Breuer
>
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>
>




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