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(_at_)star(_dot_)le(_dot_)ac(_dot_)uk]
Sent: 20 April 2005 18:32
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] 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[(_at_)name='Title_EN'] or
item[(_at_)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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