Question.. I'm using PHP 5.0.4 with the built-in libxslt-based xsl extension. I'm passing XSL parameters to the XSL processor like so: $xslt_proc->setParameter('', 'param_test', "some test value"); My question is, is it *required* to declare this parameter in the XSL stylesheet with: <xsl:param name="param_test" /> What I have noticed is, the transformation is successful and works the same whether I declare the parameter using xsl:param or not. The following: <xsl:value-of select="$param_test"/> works fine without the xsl:param declaration, the xsl processor does not give an error, and the parameter is available for use in the stylesheet. Shouldn't the xsl:param declaration be required and cause a transformation error if it's missing? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php