XSLT hyperlink replacement

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<!-- Converts any number of plain text URL strings into HTML formatted hyperlinks in a passage of text -->
<!-- Made by @george_wilson January 2012 with help from this thread - http://www.stylusstudio.com/xsllist/200305/post81170.html -->

<xsl:template name="hyperlink">
  <xsl:param name="string" select="string()" />
      <!-- Set up variables -->
<xsl:choose>
<xsl:when test="contains($string, 'http://')">
    <xsl:variable name="http" select="http://" />
</xsl:when>
<xsl:when test="contains($string, 'https://')">
    <xsl:variable name="http" select="https://" />
</xsl:when>
</xsl:choose>
      <xsl:variable name="before" select="substring-before($string, $http)" />
      <xsl:variable name="after" select="substring-after($string, 'http://')" />
      <xsl:variable name="url" select="concat($http, substring-before($after,' '))"/>
      <xsl:variable name="rest" select="substring-after($string, $url)" />
      <!-- Output the text -->
      <xsl:value-of select="$before"/>
      <xsl:choose>
              <!-- If the url is at then end, $rest doesn't work -->
      	<xsl:when test="substring-after($url,$http) != ''">
			<a href="{$url}">
				<xsl:value-of select="$url" />
			</a>
			<xsl:call-template name="hyperlink">
				<xsl:with-param name="string" select="$rest"/>
			</xsl:call-template>
		</xsl:when>
		<xsl:otherwise>
			<a href="{$url}{$after}">
				<xsl:value-of select="$after" />
			</a>
		</xsl:otherwise>
	  </xsl:choose>
</xsl:template>

</xsl:stylesheet>

I'm trying to import this handy utility in one of my Symphony installations, but I'm getting some error messages:

Invalid expression
(line 13) XSLT-variable: Failed to compile the XPath expression 'http://'.
Invalid expression
(line 16) XSLT-variable: Failed to compile the XPath expression 'https://'.

Don't know if I'm overseeing something? Hope you can push me in the right direction.