Subject: Re: [xsl] Recursive string replace in XSLT 2.0 From: "David Carlisle d.p.carlisle@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Fri, 6 Jan 2017 19:41:18 -0000 |
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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