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RE: [xsl] getting node type in xsl
Subject: RE: [xsl] getting node type in xsl From: "Michael Kay" <mhk@xxxxxxxxx> Date: Wed, 8 Sep 2004 21:59:07 +0100 |
> Still, how does xslt 2.0 without the schema aware extension help me in > my problem nr. 1? Schema-awareness isn't an "extension" to XSLT 2.0, it's an integral part of the standard. Your only problem is, it doesn't come for free :-) > So, anyone got an idea about nr. 2 (using the stylesheet to produce an > element's xpath something I can print out)? > I tried > > <xsl:template > match="node()"> > <xsl:variable name="xpath" select="concat(../$xpath, '/', > local-name(),'[', position(), ']')"/> > <fieldset> > <legend> > <xsl:value-of > select="$xpath" /> > </legend> > ... > Saxon has an extension function for this but there are templates you can pick up to do it. Google for "XSLT template to get a path to a node". Michael Kay
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