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