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Re: for-each order
- From: Trevor Nash <tcn at melvaig dot co dot uk>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Fri, 14 Dec 2001 18:37:32 +0000
- Subject: Re: [xsl] for-each order
- Organization: Melvaig Software Engineering Limited
- References: <3475147EADCBD51199150002A52C17043E3D53@MDYNYCMSX5>
- Reply-to: xsl-list at lists dot mulberrytech dot com
>Caveat: I'm a relative newbie, still weaning myself off disable output
>escaping....
>
Keep at it! ;-))
>There are a variety of ways of reversing things - just search on the archive
>for reverse, or see
>http://sources.redhat.com/ml/xsl-list/2001-08/msg00243.html
>
>but in this particular case I'm thinking that this (untested) might work:
>
> <xsl:variable name="position">
> <xsl:for-each select="$path">
> <xsl:sort order="descending" select="position()"/>
Very close. The data-type="number" is important, otherwise item 10
sorts next to item 1.
> <xsl:value-of select="@value"/>,
> </xsl:for-each>
> </xsl:variable>
>
>David.
Regards,
Trevor Nash
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