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[xsl] search and replace

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I found this template on the list but am not sure how to implement it.
I'm trying to replace ??degrees?? with ° in all xml tags. Is it
possible and can someone explain how this template would/should work?

I've tried a few variations and I keep getting an error that
call-template not allowed in stylesheet message.

Any help would be appreciated.  Thanks

<!-- sample how to use generic search-and-replace: -->
<xsl:variable name="string" select="'string to be
searched-and-replaced'"/>
<xsl:variable name="find"
select="'stringfragment-that-has-to-be-replaced'"/>
<xsl:variable name="replace"
select="'the-replacement-stringfragment'"/>
<xsl:call-template name="_replace_string">
<xsl:with-param name="string"
select="substring-after($string,$find)"/>
<xsl:with-param name="find" select="$find"/>
<xsl:with-param name="replace" select="$replace"/>
</xsl:call-template>



<!-- template to look for all instances of string $find and
replace that with string $replace -->
<xsl:template name="_replace_string">
<xsl:param name="string" select="''"/>
<xsl:param name="find" select="''"/>
<xsl:param name="replace" select="''"/>
<xsl:choose>
<xsl:when test="contains($string,$find)">
<xsl:value-of
select="concat(substring-before($string,$find),$replace)"/>
<xsl:call-template
name="_replace_string">
<xsl:with-param name="string"
select="substring-after($string,$find)"/>
<xsl:with-param name="find"
select="$find"/>
<xsl:with-param name="replace"
select="$replace"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$string"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>



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