Re: [xsl] Default values of template parameters

Subject: Re: [xsl] Default values of template parameters
From: David Carlisle <davidc@xxxxxxxxx>
Date: Tue, 1 Aug 2006 12:12:53 +0100
	<xsl:param name="linestyle" select="Solid" />

that selects the <Solid> child element of the current node. I suspect
taht you want a default of
	<xsl:param name="linestyle" select="'Solid'" />

	<xsl:variable name="UnusedParamsA" select="0,0,1,300," />
that's an XPath1 syntax error, your stylesheet shouldn't compile, I
suspect that you again want
	<xsl:variable name="UnusedParamsA" select="'0,0,1,300,'" />

   Is there a way to pass the parameter if the tested attribute exists and
   not pass anything, not even an empty string, if it does not?
only by putting the call-template inside each branch of an xsl:choose,
and passing the param in one case and not in the other.

I would however write

<xsl:with-param name="linestyle" >
		<!-- if a linestyle is specified...-->
		<xsl:when test="[@name =$MatchedEntityName]/@linestyle">
                               ^^^ again that is a syntax error
                               you can't start an xpath with [

			<!-- ... pass it's value, otherwise... -->
			<xsl:value-of select="$MatchedEntityName]/@linestyle " />
			<!-- pass nothing. -->
			<xsl:value-of select="''"/>

<xsl:with-param name="linestyle"  select=*[@name=$MatchedEntityName]/@linestyle"/>

then you are always passing $linestyle either with the correct value or
with an empty node set, so your template can go (I'm assuming you want a
* at the start of taht xpath)

 <xsl:when test="$linestyle"><xsl:value-of select="$linestyle"/></xsl:when>


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