Subject: Re: [xsl] How to substitute a portion of the text value of an element From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Tue, 07 Oct 2008 14:53:18 -0400 |
I've got an element like this:
<box:base-file-name>C:/test10/user/server/logs/server.log</box:base-file-name>
I'd like it to look like this:
<box:base-file-name>C:/other103/user/server/logs/server.log</box:base-file-name>
I'm passing in an OLD parameter that contains "c:/test10" and a NEW parameter that contains "c:/other103" but I'm not sure how to substitute just the OLD portion for the NEW parameter in the value of box:base-file-name without nuking the rest of the value which I want to preserve.
Is it possible to just change a portion of the value?
T:\ftemp>type paul.xsl <?xml version="1.0" encoding="US-ASCII"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xsd="http://www.w3.org/2001/XMLSchema" exclude-result-prefixes="xsd" version="2.0">
<xsl:template match="/"> <result> <xslt20> <xsl:apply-templates select="*" mode="x2"> <xsl:with-param name="old">C:/test10</xsl:with-param> <xsl:with-param name="new">C:/other103</xsl:with-param> </xsl:apply-templates> </xslt20> <xslt10> <xsl:apply-templates select="*" mode="x1"> <xsl:with-param name="old">C:/test10</xsl:with-param> <xsl:with-param name="new">C:/other103</xsl:with-param> </xsl:apply-templates> </xslt10> </result> </xsl:template>
<xsl:template match="*" mode="x1"> <xsl:param name="old"/> <xsl:param name="new"/> <xsl:copy> <xsl:value-of select="substring-before(.,$old)"/> <xsl:value-of select="$new"/> <xsl:value-of select="substring-after(.,$old)"/> </xsl:copy> </xsl:template>
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