Subject: Re: [xsl] Replace an attribute From: "Sean Tiley" <sean.tiley@xxxxxxxxx> Date: Fri, 4 Jan 2008 12:13:53 -0500 |
Thanks Andrew, your templates worked exactly as I wanted them to. Now I need to spend time understanding how they work. Are the expressions in the match="" statements XPath expressions? Thanks again. -- Sean Tiley sean.tiley@xxxxxxxxx When you basically want to make a copy of the input XML but with certain changes then you want to use the "indentity" transform, with specific templates to make the change. You're right to use the replace() function, but remember to get the regex right. <xsl:stylesheet xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0" exclude-result-prefixes="xs"> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="@name"> <xsl:attribute name="name"> <xsl:value-of select="replace(., 'C:\\blah\\blah', 'files')"/> </xsl:attribute> </xsl:template> </xsl:stylesheet> cheers -- Andrew Welch http://andrewjwelch.com Kernow: http://kernowforsaxon.sf.net/
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