将URL作为参数传递给XSL

Question

I'd like to pass current page URL as an attribute to XSL template. As far as I understood it should be passed as a parameter, and then used as an attribute.

I use PHP to load XML & XSL files:

<?php
$xml = new DOMDocument;
$xml->load('main.xml');

$xsl = new DOMDocument;
$xsl->load('blocks/common.xsl');

$proc = new XSLTProcessor;

$proc->importStyleSheet($xsl);

echo $proc->transformToXML($xml);
?> 

How should this code be altered to pass URL as a parameter named "current-url", for example?

I've seen a lot of similar questions here with different solutions, but none has worked for me so far. Thank you in advance.

Answer 1

Maybe you already tried this approach, but in case if not:

<?php

  $params = array('current-url' => $_SERVER['REQUEST_URI']);

  $xml = new DOMDocument;
  $xml->load('main.xml');

  $xsl = new DOMDocument;
  $xsl->load('blocks/common.xsl');

  $proc = new XSLTProcessor;
  $proc -> registerPHPFunctions();
  $proc->importStyleSheet($xsl);

  foreach ($params as $key => $val)
    $proc->setParameter('', $key, $val);

  echo $proc->transformToXML($xml);
  ?>

In the xsl, add above the templates

<xsl:param name="current-url" />

In the templates, you can get the value using

<xsl:value-of select="$current-url" />

If not already there, you have to add xmlns:php="http://php.net/xsl" into the xsl:stylesheet declaration.
For reference: registerPHPFunctions() and a solution you maybe already checked on SO: Passing variables to XSLT