Re: Recursive string replace in XSLT 2.0

I'd have written it as a function rather than template, but the main
issue is you want your parameter to be (always) a sequence of elements
not sometimes a sequence of elements and sometimes a document node
with a sequence of child elements.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="xs"
    version="2.0">

    <xsl:output indent="yes"/>

    <xsl:param name="regexes">
        <regex><find>a</find><change>x</change></regex>
        <regex><find>b</find><change>y</change></regex>
        <regex><find>c</find><change>z</change></regex>
    </xsl:param>

    <xsl:template match="/">
        <xsl:apply-templates/>
    </xsl:template>

    <xsl:template match="p">
        <p><xsl:apply-templates/></p>
    </xsl:template>

    <xsl:template match="text()"><!--[string-length(.)>0]-->
        <xsl:message select="."></xsl:message>
        <xsl:call-template name="applyRegexes">
            <xsl:with-param name="nodeText" select="."/>
            <xsl:with-param name="regex" select="$regexes/regex"/>
         </xsl:call-template>
    </xsl:template>

    <xsl:template name="applyRegexes">
        <xsl:param name="nodeText"/>
        <xsl:param name="regex"/>
        <xsl:message select="$regex"></xsl:message>
        <xsl:message select="$regex[1]"/>
        <xsl:message select="$regex[position()>1]"/>
        <xsl:choose>
            <xsl:when test="$regex">
                <xsl:call-template name="applyRegexes">
                    <xsl:with-param name="nodeText"
select="replace($nodeText,$regex[1]/find,$regex[1]/change)"/>
                    <xsl:with-param name="regex"
select="$regex[position()>1]"/>
                </xsl:call-template>
            </xsl:when>
            <xsl:otherwise>
                <xsl:value-of select="$nodeText"/>
            </xsl:otherwise>
        </xsl:choose>
     </xsl:template>

</xsl:stylesheet>