Hi
-----Original Message-----
From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
[mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of
Tengshe, Ashish
Sent: Wednesday, June 25, 2003 7:40 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: RE: [xsl] No to display the values which are already existing
Thanks, Works fine Now!
You could also use here the muenchian method, something like:
<xsl:key match="Title" name="list" use="Author"/>
<xsl:template match="Store">
<table>
<tr>
<td>Author</td>
<td>Title</td>
</tr>
<xsl:apply-templates mode="header"
select="Title[generate-id()=generate-id(key('list',Author))]"/>
</table>
</xsl:template>
<xsl:template match="Title" mode="header">
<tr>
<td><xsl:apply-templates select="Author"/></td>
<td><xsl:apply-templates select="Name"/></td>
</tr>
<!-- don't need to select the current one because it already have been
displayied -->
<xsl:apply-templates
select="key('list',Author)[not(generate-id()=generate-id(current()))]"/>
</xsl:template>
<xsl:template match="Title">
<tr>
<td><xsl:text> </xsl:text></td>
<td><xsl:apply-templates select="Name"/></td>
</tr>
</xsl:template>
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list