AW: [xsl] Selecting out groups for a menu

["Zink, Juergen" <Juergen.Zink@xxxxxxxxxxxxxxxxxxxx>]

Hello John,

<snip>
>The first group contains all the pages that have no children and sorts them
in order by there priority the following groups >list by first the parent
group as the title and then the childrens below and list them by priority.
Unfortunatly this is all >over my head. Im trying my best to learn the
tutorials and figure out how i can do this but I am very stuck on how to
group >this like above. Thanks for any help or info you can give me -john

Select the children that have no children (first apply-templates) with
sorting
Select the children that have children (second apply-templates)
list by priority (third apply-templates)

Assuming there will be ever Pages with no children.

  <xsl:template match="/">
    <xsl:apply-templates select="PageHierarchy/Page[not(Page)]">
      <xsl:sort select="@priority" />
    </xsl:apply-templates>
    <xsl:apply-templates select="PageHierarchy/Page[Page]">
      <xsl:sort select="@priority" />
    </xsl:apply-templates>
  </xsl:template>

  <xsl:template match="Page">
    <xsl:if test="not(parent::Page) and Page">
      <xsl:text>--------space----</xsl:text>
      <xsl:element name="br"/>
    </xsl:if>
    <xsl:value-of select="@actualname"/>
    <xsl:element name="br"/>
    <xsl:apply-templates select="Page">
      <xsl:sort select="@priority" />
    </xsl:apply-templates>
  </xsl:template>

Kind regards,

Juergen Zink


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