Re: [xsl] re-ordering nodes

[Jeni Tennison <jeni at jenitennison dot com>]

David,

> Jeni> Perhaps you can make it clearer what you want the stylesheet
> Jeni> to do.
>
> Jeni, I'm disapointed in you, you used to be able to reliably
> determine the spec given no information at all, have you lost this
> gift?

Actually I was being lazy. I think that what Srini wanted was:

<xsl:strip-space elements="*" />

<xsl:template match="list">
  <xsl:apply-templates select=".//o">
    <xsl:sort select="count(node())" data-type="number" />
  </xsl:apply-templates>
</xsl:template>

<xsl:template match="o">
  <o>
    <xsl:apply-templates select="a | b | text()" />
  </o>
</xsl:template>

<xsl:template match="a | b">
  <xsl:copy>
    <xsl:copy-of select="@*" />
    <xsl:if test="o">
      <xsl:comment>
        <xsl:text> for </xsl:text>
        <xsl:for-each select=".//o[not(o)]">
          <xsl:value-of select="number()" />
          <xsl:if test="position() != last()">, </xsl:if>
        </xsl:for-each>
        <xsl:text> </xsl:text>
      </xsl:comment>
    </xsl:if>
  </xsl:copy>
</xsl:template>

At least it gives the required output for the single sample
document...

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


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