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Damon BetlowSubject: Wildcard Param
Author: Damon Betlow
Date: 12 May 2008 01:00 PM
Hello,

How does one use a wildcard as Param value? It appears that the '*' is being treated as text instead of a wildcard.

If my PHP page does not pass in a 'category param', I want the XSLT file to transform all of the 'blog postings'. If I pass in a category, it works fine. I also tried changing it so that I could pass in either '/blog/post' or '/blog/post[category="whatever"]', but that didn't seem to work either.

Thanks in advance!

XSLT file (relevant code):

<xsl:param name="category" select="*" />

<xsl:template match="/">
<xsl:apply-templates select="/blog/post[category=$category]">
<xsl:sort select="position()" order="descending" />
</xsl:apply-templates>
</xsl:template>

PHP file (relevant code):

/* create the processor and import the stylesheet */
$proc = new XsltProcessor();
$xsl = $proc->importStylesheet($xsl);
$proc->setParameter(null, "category", "Family");
Postnext
Tony LavinioSubject: Wildcard Param
Author: Tony Lavinio
Date: 12 May 2008 02:52 PM
Please ask general XSLT questions on the xsl-list run by
Mulberry Technologies.

This list is for XSLT questions having to do with Stylus
Studio, as part of the Stylus Studio Developer Network.
Posttop
James DurningSubject: Wildcard Param
Author: James Durning
Date: 20 May 2008 04:11 PM
Add an extra apply-templates clause for this special instance.

<xsl:param name="category" select="*" />

<xsl:template match="/">
<xsl:apply-templates select="/blog/post[category=$category]">
<xsl:sort select="position()" order="descending" />
<xsl:if test="$category='*'"><!-- wildcard -->
<xsl:apply-templates select="/blog/post">
</xsl:if>
<xsl:sort select="position()" order="descending" />
</xsl:apply-templates>
</xsl:template>
   
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