xsl：param默认值未被覆盖

Question

I have tried reducing my problem to as small a test case as possible.

I currently have the following stylesheet:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:template match="node()|@*">
      <xsl:copy>
         <xsl:apply-templates select="node()|@*"/>
      </xsl:copy>
   </xsl:template>

   <xsl:template name="set-attrib">
      <xsl:param name="val" select="3"/>
      <xsl:attribute name="value">
         <xsl:value-of select="$val" />
      </xsl:attribute>
   </xsl:template>

   <xsl:template match='@value[parent::element]'>
      <xsl:call-template name="set-attrib">
         <with-param name="val">2</with-param>
      </xsl:call-template>
   </xsl:template>

</xsl:stylesheet>

And applying it to the following XML:

<?xml version="1.0"?>
<element value="0"/>

I get the output:

<?xml version="1.0"?>
<element value="3"/>

rather than what I'd expect:

<?xml version="1.0"?>
<element value="2"/>

It appears that the template set-attrib is ignoring the parameter val that is being set in call-template .

I have dug around the net for ideas, and tried varied permutations (eg using select="2" in the with-param , or moving the default value of xsl:param into the <!-- Content: template --> , rather than in the select attribute) with no difference.

The only way I seem to be able to get it to work is by hard-coding the parameter within the set-attrib template or completely inlining the call-template .

I cannot see how this is different to other example code; am I missing something blatantly obvious?

Answer 1

<with-param name="val">2</with-param> needs to be <xsl:with-param name="val">2</with-param> . With that change, http://xsltransform.net/bFDb2BS works fine.