dl/dt/dd matching

[Francois Belanger <francois@xxxxxxxxxxx>]

I'm scratching my head on this one. We are trying to define an XML 
content DTD and giving a shot to Voyager, the W3 effort for reinstating 
HTML 4 as an XML application see <http://www.w3.org/TR/WD-html-in-xml/>.

What were trying to do is layout a suite of dt/dd within a dl as a table 
with a row for each dt. Since there's no container for each dt (the 
corresponding dl are siblings, not children of the dt node), how can one 
start a new row for each dt and express that in XSL?

Speicifcally, any suggestion on xsl:templates for turning:

<dl>
<dt>Term 1</dt>
<dd>Description 1</dd>
<dt>Term 2</dt>
<dd>Description 2</dd>
</dl>

into:

<TABLE>
<TR>
     <TD>Term1</TD>
     <TD>Description 1</TD>
</TR>
<TR>
     <TD>Term2</TD>
     <TD>Description 2</TD>
</TR>
</TABLE>

Our current templates (which obviously do not produce the desired result):

<xsl:template match="dl">
	<TABLE>
	<xsl:apply-templates/>
	</TABLE>
</xsl:template>

<xsl:template match="dl/dt">
	<TR>
    <TD><xsl:value-of select="."/></TD>
		<TD<xsl:apply-templates select = "../dd"></TD>
   </TR>
</xsl:template>

<xsl:template match="dl/dd">
	<TD><xsl:apply-templates/></TD>
</xsl:template>

Any suggestion or comment welcome!

Francois Belanger
Sitepak, Bringing Internet Business into Focus
http://www.sitepak.com



 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list