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Re: Pass-Through of MathML-Elements
- From: Jeni Tennison <jeni at jenitennison dot com>
- To: Maik Stührenberg <Maik dot Stuehrenberg at germanistik dot uni-giessen dot de>
- Cc: XSL-List at lists dot mulberrytech dot com
- Date: Tue, 14 May 2002 11:36:11 +0100
- Subject: Re: [xsl] Pass-Through of MathML-Elements
- Organization: Jeni Tennison Consulting Ltd
- References: <5.1.0.14.2.20020514114429.00bfde98@popg.uni-giessen.de>
- Reply-to: xsl-list at lists dot mulberrytech dot com
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/
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