[xsl] Re: Re: Bug in treating an RTF by Saxon 6.5 and MSXML

Subject: [xsl] Re: Re: Bug in treating an RTF by Saxon 6.5 and MSXML
From: Dimitre Novatchev <dnovatchev@xxxxxxxxx>
Date: Thu, 16 May 2002 01:21:12 -0700 (PDT)
"Michael Leditschke" <mike at ammd dot com dot au> wrote:

> If the select attribute is used on a xsl:param, isn't the value of
> the parameter the result of the XPath expression, in which case its 
> a boolean, a number, a string, or a nodeset? Its only an RTF is you
> include a template in the body of the xsl:param elements. So the
> correctness or otherwise of the later operations depends on what
> XPath expression you use.


In the example provided, the template that receives the RTF as the
value of its $arg1 parameter is:

    <xsl:template match="str-split2words-func:*">
      <xsl:param name="arg1" select="/.."/>
      <xsl:param name="arg2"/>
......................Some nodes are copied here to the result tree

It is instantiated in the (provided) stylesheet str-foldl.xsl in the
following way:

            <xsl:variable name="vFunResult">
              <xsl:apply-templates select="$pFunc[1]">
                <xsl:with-param name="arg0" 
                                select="$pFunc[position() > 1]"/>
                <xsl:with-param name="arg1" select="$pA0"/>
                <xsl:with-param name="arg2" 

            <xsl:call-template name="str-foldl">
		<xsl:with-param name="pFunc" select="$pFunc"/>
		<xsl:with-param name="pStr" 
		<xsl:with-param name="pA0" select="$vFunResult"/>

First a new temporary result is accumulated in $vFunResult. This is an
RTF, which has been produced by applying the first template above.

So in the xsl:call-template, the str-foldl template calls itself
recursively, passing this RTF as the value of the pA0 parameter, which
then is passed as the value of the $arg1 parameter, when the variable
$vFunResult is calculated again.

So, $arg1 contains an RTF.

Dimitre Novatchev.

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