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Re: [xsl] Default values of template parameters

2006-08-01 04:13:19


        <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" >
        <xsl:choose>
                <!-- if a linestyle is specified...-->
                <xsl:when test="[(_at_)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 " 
/>
                </xsl:when>
                <xsl:otherwise>
                        <!-- pass nothing. -->
                        <xsl:value-of select="''"/>
                </xsl:otherwise>
        </xsl:choose>
</xsl:with-param>

as
<xsl:with-param name="linestyle"  
select=*[(_at_)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:choose>
 <xsl:when test="$linestyle"><xsl:value-of select="$linestyle"/></xsl:when>
 <xsl:otherwise>Solid</xsl:otherwsie>

David





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