PHP and XSLTProcessor Misbehavior

Question

Simple question: Why is a PHP function called from an XSL Stylesheet just returning the last argument passed:

foo.xsl:

<xsl:template match="/">
    <xsl:value-of select="php:function('date','c')" />
</xsl:template>

PHP:

...  
$xsl = new XSLTProcessor();
$xsl->registerPHPFunctions();
$xsl->importStylesheet($fooStylesheet);
echo $xsl->transformToXML($myXML);

I Get the output

and if I call <xsl:value-of select="php:function('date')" /> I just get date as my output. Seems strange to me.

Version info:
PHP 5.3.2
libxslt Version 1.1.26
libxslt compiled against libxml Version 2.7.6
EXSLT enabled
libexslt Version 1.1.26

Did you put the PHP namespace in your stylesheet root tag? Like `` — zneak, May 29 '10 at 18:34

score 1 · Accepted Answer · answered May 29 '10 at 18:41

1

You need the PHP xmlns in your <stylesheet> root tag:

<stylesheet xmlns:php="http://php.net/xsl">

answered May 29 '10 at 18:41

zneak

Fairly straight forward, but I had been starting at that stylesheet for an hour! – Aiden Bell May 29 '10 at 18:53

1 Answers1