Excellent David! Thank you for your generous help. -Rick
-----Original Message-----
From: David Carlisle d(_dot_)p(_dot_)carlisle(_at_)gmail(_dot_)com
[mailto:xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com]
Sent: Friday, January 06, 2017 2:42 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] 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>
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