Alternative search and replace template for XSLT 1.0.

search-and-replace-2.xsl

<!--
    ALTERNATIVE SEARCH & REPLACE
    string:     The text to be evaluated
    replace:    The character or string to look for in the above string
    with:       What to replace it with

    Slightly more long winded approach if that's how you prefer to roll.
--> 

<xsl:template name="string-replace">
    <xsl:param name="string" />
    <xsl:param name="replace" />
    <xsl:param name="with" />

    <xsl:choose>
        <xsl:when test="contains($string, $replace)">
            <xsl:value-of select="substring-before($string, $replace)" />
            <xsl:value-of select="$with" />
            <xsl:call-template name="string-replace">
                <xsl:with-param name="string" select="substring-after($string,$replace)" />
                <xsl:with-param name="replace" select="$replace" />
                <xsl:with-param name="with" select="$with" />
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:value-of select="$string" />
        </xsl:otherwise>
    </xsl:choose>
</xsl:template>

<!-- Example template call -->
<xsl:variable name="string-mod">
    <xsl:call-template name="string-replace">
        <xsl:with-param name="string" select="'This is a sample text : {replace} and {replace}'" />
        <xsl:with-param name="replace" select="'{replace}'" />
        <xsl:with-param name="with" select="'{I've been replaced}" />
    </xsl:call-template>
</xsl:variable>

<!-- Output -->
<xsl:value-of select="string-mod" />

<!-- Result: "This is a sample text : {I've been replaced} and {I've been replaced}" -->

The with param tag is missing a [']. It should be: <xsl:with-param name="with" select="'{I've been replaced}'" />