Subject: [xsl] Sorting, but then splitting into 5 <ul> columns From: "Dan Acuff" <dacuff@xxxxxxxxxxxxxx> Date: Wed, 6 Aug 2008 09:35:54 -0400 |
I have figured out how to sort based on an attribute value. But so far I can only get 1 UL list, and I really need to use the columnCount variable to make 5 <UL> lists, that we will then float left. Is there a way to do this possibly by using the results of the first category template and processing it in a new template? <xsl:param name="paraCatagory"/> <xsl:param name="altRowBGColor"/> <xsl:template match="menu"> <xsl:apply-templates select="descendant::category[@name=$paraCatagory]"/> </xsl:template> <xsl:template match="category"> <xsl:variable name="modelCount" select="count(category)"/> <xsl:variable name="columnCount" select="round($modelCount div 5) +1"/> <ul> <xsl:for-each select="child::category"> <xsl:sort data-type="text" order="ascending" select="@display_name"/> <xsl:choose> <xsl:when test="position() mod 2 = 0"> <li style="background-color:{$altRowBGColor};"><!-- Set this color in includes/code/category_vb.asp --> <a href="{descendant::link}"> <xsl:value-of select="@display_name"/> </a> </li> </xsl:when> <xsl:otherwise> <li> <a href="{descendant::link}"> <xsl:value-of select="@display_name"/> </a> </li> </xsl:otherwise> </xsl:choose> </xsl:for-each> </ul> </xsl:template> </xsl:stylesheet> Thank you, Dan ~~~~~~~~~~~~~~~~~~ Dan Acuff Developer SureSource, Inc. 203.922.7546 dacuff@xxxxxxxxxxxxxx ~~~~~~~~~~~~~~~~~~
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