[xsl] question about transforming columns into dropdown list?

Subject: [xsl] question about transforming columns into dropdown list?
From: "dmitri kerievsky" <dmitrik@xxxxxxxxxxxxxx>
Date: Wed, 12 Jun 2002 12:05:43 -0400
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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