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xsl:sort -- XSLT version 1.1/2.0: document order and request for reverse doc. order
- To: xsl-list at lists dot mulberrytech dot com
- Subject: [xsl] xsl:sort -- XSLT version 1.1/2.0: document order and request for reverse doc. order
- From: vermet at cs dot utwente dot nl (Eric Vermetten)
- Date: Thu, 3 May 2001 22:52:26 +0200 (MET DST)
- Cc: vermet at cs dot utwente dot nl
- Reply-To: xsl-list at lists dot mulberrytech dot com
XML elements that are already in some kind of order
in the source document are processed in order they are written in,
this order is refferred to as document order. Since WD XSLT v1.1 it is
stated explicitly that when such elements/nodes are used in the
select att of an apply-templates/xsl:for-each --without an xsl:sort as their child--
they are processed in document order (XSLTv1.1 ch. 10).
Sometimes, however, I want to process these nodes in
*reverse* order.
Q: Is there some deaper reasoning why such a possibility isn't offered?
Suggestion:
Would it be a usable idea to extent the order att of the xsl:sort in the
following manner:
order [optional]
values:
ascending
descending
document-order
reverse-document-order
The default remaining ascending.
On the minus side we now have some redundancy:
<xsl:for-each select="test">
...
</xsl:for-each>
would now have the same semantics as:
<xsl:for-each select="test">
<xsl:sort order="document-order" />
...
</xsl:for-each>
Any ideas?
thanks,
Eric
P.S. I realise that such a reverse-document order can be programmed
with the node-set extension, but that doesn't seem a very concise solution.
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