Subject: [xsl] Template match via xsl:param From: Mario Lang <mlang@xxxxxxxxxxx> Date: 14 Jan 2002 22:03:40 +0100 |
Hello. I am trying to write some generic stylesheets which I can stack in filterproxy. >From the archive/FAQ I figure that this isnt as easy as one may think, but I couldnt find a simple example which illustrates how to do it right. So, here is what I am trying to do: <?xml version="1.0"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="html" indent="yes" encoding="iso8859-15"/> <xsl:param name="path"></xsl:param> <xsl:include href="identity.xsl"/> <xsl:template match="$path"/> </xsl:stylesheet> Can anyone give me a simple example on how to achieve this without genereting the xsl file on the fly? -- CYa, Mario <mlang@xxxxxxxxxxx> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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