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> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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