[xsl] RE: translate quote into quote quote

I fixed it :

<xsl:template name="escapeQuotes">
		<xsl:param name="string"/>
		<xsl:variable name="apos">'</xsl:variable>
		<xsl:if test="not(contains($string,$apos))"><xsl:value-of
select="$string"/></xsl:if>
		<xsl:if test="contains($string,$apos)"><xsl:value-of
select="substring-before($string,$apos)"/><xsl:value-of
select="$apos"/><xsl:value-of select="$apos"/><xsl:call-template
name="escapeQuotes"><xsl:with-param name="string"
select="substring-after($string,$apos)"/></xsl:call-template></xsl:if>
	</xsl:template>

Philippe


-----Message d'origine-----
De : Philippe LAPLANCHE
Envoyi : lundi 12 dicembre 2005 07:23
@ : xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Objet : translate quote into quote quote

Hi,
I need to translate a string such as this one :
I'm sad that I can't do it by myself
Into :
I''m sad that I can''t do it by myself

I'm using XSLT 1.0, and I tried to do something like that :

<xsl:template name="escapeQuotes">
		<xsl:param name="string"/>
		<xsl:variable name="apos" select="'"/>
		<xsl:if test="not(contains($string,$apos))"><xsl:value-of
select="$string"/></xsl:if>
		<xsl:if test="contains($string,$apos)">
			<xsl:value-of select="substring-before($string,$apos)"/><xsl:value-of
select="$apos"/><xsl:value-of select="$apos"/><xsl:call-template
name="escapeQuotes"><xsl:with-param
name="substring-after($string,$apos)"/></xsl:call-template>
		</xsl:if>
	</xsl:template>


I'm doing something wrong ...

Philippe