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question about transforming columns into dropdown list?
- From: "dmitri kerievsky" <dmitrik at mindspring dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Wed, 12 Jun 2002 12:05:43 -0400
- Subject: [xsl] question about transforming columns into dropdown list?
- Reply-to: xsl-list at lists dot mulberrytech dot com
this code produces hyperlinked values in columns. What is required to place
those same values into a drop down list?
tia
<!-- Named templates
-->
- <xsl:template name="albumColumns">
<xsl:param name="column" select="1" />
<xsl:param name="startPosition" select="1" />
<xsl:param name="albums" />
<xsl:param name="columnCount" />
<xsl:param name="albumsPerColumn" />
- <td valign="top" width="{100 div $columnCount}%">
<xsl:apply-templates select="$albums/title[position() >=
$startPosition][position() <= $albumsPerColumn]" mode="AlbumDetailLink" />
</td>
- <!-- recurse
-->
- <xsl:if test="$column < $columnCount">
- <xsl:call-template name="albumColumns">
<xsl:with-param name="column" select="$column + 1" />
<xsl:with-param name="startPosition" select="$startPosition +
$albumsPerColumn" />
<xsl:with-param name="albums" select="$albums" />
<xsl:with-param name="columnCount" select="$columnCount" />
<xsl:with-param name="albumsPerColumn" select="$albumsPerColumn" />
</xsl:call-template>
</xsl:if>
</xsl:template>
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