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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] dl/dt/dd matching
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
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