Subject: [xsl] xsl:sort From: Phillip Nicolson <pjn3@xxxxxxxxxxxxx> Date: 20 Apr 2005 18:32:13 +0100 |
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. > > __________________________________________________ > Do You Yahoo!? > Tired of spam? Yahoo! Mail has the best spam protection around > http://mail.yahoo.com
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