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Re: Transformation from XML to Table Rows
- To: <xsl-list at mulberrytech dot com>
- Subject: Re: Transformation from XML to Table Rows
- From: Estefan27 at aol dot com
- Date: Mon, 04 Dec 2000 19:26:36 EST
- Reply-To: xsl-list at mulberrytech dot com
never mind i figured it out
<BODY bgcolor="white" text="#666666" >
<table width="700px" border="0" cellspacing="0" cellpadding="0" align="center">
<tr>
<xsl:for-each select="channel_list/topic">
<td align="left" valign="absbottom"><span class="channelname">
<xsl:element name="a">
<xsl:attribute name="href">
/channel.asp?catid=<xsl:value-of select="@id"/>
</xsl:attribute>
<xsl:value-of select="topic_name"/>
<hr noshade="noshade" color="#EBEBEB" align="left" width="200px" />
</xsl:element>
<span class="channeltopic">
<xsl:apply-templates select="sub_topic" ></xsl:apply-templates >
</span>
<xsl:if test="position() mod 3 = 0">
<xsl:element name="tr">
<xsl:attribute name="height">25px
</xsl:attribute>
</xsl:element>
</xsl:if>
</span></td>
</xsl:for-each>
</tr>
</table>
</BODY>
</HTML>
</xsl:template>
<xsl:template match="sub_topic" >
<xsl:element name="a">
<xsl:attribute name="href">
/channel.asp?catid=<xsl:value-of select="@id"/>
</xsl:attribute>
<xsl:value-of select="sub_topic" />
<xsl:apply-templates />
</xsl:element>
<br />
</xsl:template>
</xsl:stylesheet>
In a message dated Mon, 4 Dec 2000 7:15:27 PM Eastern Standard Time, David Carlisle <davidc@nag.co.uk> writes:
<<
Your example document had a document element called root with
child elements called topic, but your XSL template for
/ has a for-each selecting
<xsl:for-each select="channel_list/topic">
which won't select anything as the child elemnt of / is called root
not channel_list.
also your root template never calls apply-templates so the other
templates in the styplesheet that you define can never possibly have any
effect.
David
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