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Re: reordering output (Again)
- To: Eric Smith <Eric dot Smith at fruitcom dot com>
- Subject: Re: [xsl] reordering output (Again)
- From: Jeni Tennison <jeni at jenitennison dot com>
- Date: Mon, 22 Oct 2001 14:34:03 +0100
- CC: xsl-list at lists dot mulberrytech dot com
- Organization: Jeni Tennison Consulting Ltd
- References: <20011022144229.F22646@apple.fruitcom.com><20011022150206.B11166@dragon.blacknet.de>
- Reply-To: xsl-list at lists dot mulberrytech dot com
Goetz Block wrote:
> On Mon, Oct 22 '01 at 14:42, Eric Smith wrote:
>> In short, perhaps someone may advise:
>> Is reordering of nodes in the output possible?
> Yes.
>
> But could you be more elaborated on what and how you want to
> reorder. If you have a specific criteria, you can use <xsl:for-each
> sort=''>, but it might be more complicated than this.
Goetz meant:
<xsl:for-each select="...">
<xsl:sort select="..." />
...
</xsl:for-each>
You can also use xsl:sort within xsl:apply-templates to change the
order in which nodes are processed with templates.
I hope that helps,
Jeni
---
Jeni Tennison
http://www.jenitennison.com/
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