[xsl] xsl:sort

[Phillip Nicolson <pjn3@xxxxxxxxxxxxx>]

Hi,

I am trying to sort a list of jobs using:

            <xsl:for-each select="//job">
            <xsl:sort select="@time" />
            ........etc

However the value of time is sorted by the day not date order ie:

Fri Apr 01 14:55:08 BST 2005

will come before 

Thu Mar 31 13:54:55 BST 2005

as the sort is using the default data-type and sorting as text. Is there
a way to sort these values using the date values of @time?

thanks



On Wed, 2005-04-20 at 18:17, Pierre-Yves wrote:
> Hello,
> 
> Thanks for the answers but it's not so easy :
> 
> I gave Paragraph, Title_EN and SubParagraph_EN as examples but I have
> many items that have other names.
> 
> doing something like item[@name='Title_EN'] or item[@name='Paragraph']
> won't help me much...
> 
> I am trying with things like :
> <item name="{substring-before(concat(@name, '_'), '_')}">
> to get all items names without underscore something at the end
> and 
> [substring-after(@name, '_') to get EN or FR
> but still I don't succeed in making something that outputs what I
> expect.
> 
> Thanks,
> Pierre.
> 
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