RE: xsl:sort

You will need to convert the dates/times to standard ISO 8601 notation

2005-04-01T14:55:08+01:00

Code for doing such conversion has been posted on this list last week.

Michael Kay
http://www.saxonica.com/ 

> -----Original Message-----
> From: Phillip Nicolson [mailto:pjn3@xxxxxxxxxxxxx] 
> Sent: 20 April 2005 18:32
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject:  xsl:sort
> 
> 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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