[xsl] Re: xsl:with-param

> Am I misunderstanding something or should the follow xslt
> snippet work:
>
> -----
> <xsl:template match="/">
> <foo>
>    <xsl:apply-templates mode="foo">
>     <xsl:with-param name="wit">B</xsl:with-param>
>   </xsl:apply-templates>
> </foo>
> </xsl:template>
>
> <xsl:template match="body" mode="foo">
>   <xsl:param name="wit" />
>   <xsl:if test="contains(@wit, $wit)">
> <body wit="{$wit}"><xsl:apply-templates mode="foo"/></body>
> </xsl:if>
> </xsl:template>
> -----
>
> I expected that the root template would apply any
> other templates with mode="foo" and make available to
> them the param $wit with value 'B'.

In case "body" is not the top element (the child of "/"), probably a
template that was instantiated for one of its ancestors didn't have an
xsl:param named "wit".

Then "the chain is broken" ... this template cannot pass a parameter it
doesn't know about.

You haven't shown us the sml.source and the other templates that are part of
the transformation, so one can only provide a guess of what is the exact
reason for the reported behaviour.



=====
Cheers,

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





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