[xsl] Sorting and re-ordering down a hierarchy

Subject: [xsl] Sorting and re-ordering down a hierarchy
From: "Cole, Chris" <chris.cole@xxxxxxxxxxxxxxxxxxxxx>
Date: Fri, 18 Jul 2003 15:00:30 -0000
Hello,
 I'm struggling to re-organise my node tree.
 My tree is something like this:

<input>
<node>
    <rank>2</rank>
    <node>
        <rank>88</rank>
    </node>
    <node>
        <rank>7</rank>
    </node>
    <node>
        <rank>66</rank>
    </node>
</node>
<node>
    <rank>1</rank>
    <node>
        <rank>3</rank>
    </node>
    <node>
        <rank>2</rank>
    </node>
    <node>
        <rank>1</rank>
    </node>
</node>
</input>

I need to reorganise the nodes so that they are organised in number order
down the nodes within their tree, so the output should be like this:
<output>
<node>
    <rank>1</rank>
    <node>
        <rank>7</rank>
    </node>
    <node>
        <rank>66</rank>
    </node>
    <node>
        <rank>88</rank>
    </node>
</node>
<node>
    <rank>2</rank>
    <node>
        <rank>1</rank>
    </node>
    <node>
        <rank>2</rank>
    </node>
    <node>
        <rank>3</rank>
    </node>
</node>
</output>
I can sort one level by using the code below, but I'm struggling to find a
way to re-order the second level within the first level (and the third level
later).

<xsl:template match="/input">
  <xsl:call-template name="ascending-numeric-sort-l1">
    <xsl:with-param name="seqnuml1" select="node/rank"/>
  </xsl:call-template>

</xsl:template>

<xsl:template name="ascending-numeric-sort-l1">
  <xsl:param name="seqnuml1"/>
  <xsl:for-each select="$seqnuml1">
    <xsl:sort select="." data-type="number" order="ascending"/>
    <xsl:copy-of select="parent::node()"/>
  </xsl:for-each>
</xsl:template>

Can anybody offer any advice please?
Thanks
Chris

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