Subject: Re: [xsl] Pass-Through of MathML-Elements From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> Date: Tue, 14 May 2002 11:36:11 +0100 |
Hi Maik, > I thought the default rule for XSLT processors is to pass the output > through so I deleted the whole Formula template - same result. What > output parameter, encoding or template do I need to define to get > MathML elements pass through? The default behaviour, if you don't have templates for elements, is to get their *text content*. This is exactly the same as getting their string value -- all of their text descendants concatenated together -- which is what you get when you use xsl:value-of. You should be using xsl:copy-of to copy the structure. Try: <xsl:template match="Formula"> <xsl:choose> <xsl:when test="@Display='Block'"> <blockquote> <xsl:copy-of select="node()"/> </blockquote> </xsl:when> <xsl:otherwise> <xsl:copy-of select="node()"/> </xsl:otherwise> </xsl:choose> </xsl:template> Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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