How to pass parameter to xsl using xslt processor?

Question

I am getting an empty result when trying to pass a parameter to my stylsheet using the php xslt processor. Any ideas would be greatly appreciated!

PHP:

<?php 
$xsl = new DomDocument();
$xsl->load("style_listgastro.xsl");
$inputdom = new DomDocument();
$inputdom->load("XXX");
$proc = new XsltProcessor();
$xsl = $proc->importStylesheet($xsl);
$xsl = $proc->setParameter(null, 'k', $_GET['k']);
$newdom = $proc->transformToDoc($inputdom);
print $newdom->saveXML();
?>

XSLT:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"     xmlns:td="urn:schemas-XXX:Relationen">
<xsl:param name="k" />
<xsl:template match="/td:Relationen">
...
<xsl:apply-templates select="td:Benutzer_zu_Gastro">
<xsl:sort select="td:Gastroname"></xsl:sort>
</xsl:apply-templates>
...
</xsl:template>
<xsl:template match="td:Benutzer_zu_Gastro">
<xsl:if test="td:Kategorien = '{$k}'">
<li><a>
<xsl:attribute name="href">XXX</xsl:value-of>
</xsl:attribute>
<xsl:value-of select="td:Gastroname"></xsl:value-of>
</a></li>
</xsl:if>
</xsl:template>
</xsl:stylesheet>

JLRishe · Accepted Answer · 2013-03-11 11:09:34Z

2

This line is incorrect:

<xsl:if test="td:Kategorien = '{$k}'">

it's not comparing td:Kategorien against the value of $k, but against the string value {$k} (and I'm guessing that td:Kategorien will never have the value {$k}). Instead, just use this:

<xsl:if test="td:Kategorien = $k">

Or, a better fix would be to change the apply-templates line to this:

<xsl:apply-templates select="td:Benutzer_zu_Gastro[td:Kategorien = $k]">

and remove the xsl:if altogether.

edited Mar 11, 2013 at 11:09

answered Mar 11, 2013 at 10:58

JLRishe

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Comments

djc · Accepted Answer · 2013-03-11 09:56:39Z

0

Try using '' instead of null as the first argument for setParameter()?

answered Mar 11, 2013 at 9:56

djc

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