if you sort by the reverse of your date(26.11.2003) i.e. 20031126 ....
you can extract and sort using substring on your date i.e.
<xsl:foreach.............
<xsl:sort data-type="number" select="substring(@date,6,4)"/>/<xsl:value-of
select="substring(@date,3,2)"/>/<xsl:value-of select="substring(@date,1,2)"
order="ascending"/>
<xsl:if position()=1>
</xsl:if>
Syntax and substring result is most likely wrong but i think it would work.
PA Sport RnD
Andrew Curry
Software Developer
Telephone: 01430 455545
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----- Original Message -----
From: "Iris Benjamin" <Benjamin(_dot_)Iris(_at_)bwg1(_dot_)siemens(_dot_)de>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Friday, December 05, 2003 10:16 AM
Subject: [xsl] Date variable ?!
Hello
I have a simple XML document wit lots of elements : "testCase"
every "testCase" has an attribute such as startedOn="26.11.2003
11:41:30"
In my xslt programm, I would like to extract the earliest startedOn
attribute among all the testCase elements.
That means, look through all the values of the attribute and find a way
tget
the earliest date.
(if it is too complicated with the Time, we can only focus on the date)
If you have any idea, I thank you in advance.
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list