RE: Re: [xsl] XSLT 2.0/XPath 2.0 Date arithmetic
Subject: RE: Re: [xsl] XSLT 2.0/XPath 2.0 Date arithmetic|
Date: Wed, 17 May 2006 21:06:07 -0400
That's the direction I was going, but I suppose I was getting tangled in the substring()s and concat() that I was using to produce the XPath 2.0 (or is it XSLT 2.0) format for a date. I was staring at the very volumes you mention, but I'll take another crack at it tomorrow when I'm fresh.
cknell@xxxxxxxxxx - email
From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>
Sent: Wed, 17 May 2006 18:33:05 -0400
Subject: Re: [xsl] XSLT 2.0/XPath 2.0 Date arithmetic
The new time-and-date data types work very well for this stuff. First
contrive a way to cast your date string to an honest date value (this
will involve munging its format a bit to the normative 'YYYY-MM-DD'
arrangement, then casting it). At that point you can simply
"subtract" days from it. Date arithmetic allows the addition and
subtraction of dayTimeDuration values from dates (and times): very handy.
Here's a snippet of code that does something not dissimilar:
<xsl:variable name="day" select="xdt:dayTimeDuration('P1D')"/>
select="(for $d in (0 to 6) return
($today - ($d * $day)))[format-date(.,'[F]')='Monday']"/>
Given a date value $today, $thisMonday is bound to the
immediately-preceding Monday. That is, if $today is (a date value)
'2006-05-17', the select expression here iterates over $today and the
immediately-preceding six days, and returns those members of the
sequence that turn out to be Mondays (which happens to be
'2006-05-15'; but if $today were 2006-06-01, $thisMonday would be 2006-05-29).
The short version: if $today = xs:date($string) (and $string is in
the form of "2006-05-17"), then
$tMinus1 = ($today - xdt:dayTimeDuration('P1D'))
I'm finding both Mike Kay's XPath 2.0 book, and Jeni Tennison's
Learning XSLT 2.0 book, to be invaluable for figuring all this out.
Honestly, I don't know what we'd do without them.
At 05:29 PM 5/17/2006, you wrote:
Given a parameter, let's call it "today" in the form of this string
"20060517", how do I create a variable, let's call it "tMinus1" such
that it represents a day earlier than "20060517", that would be
"20060516". So long as "$today" isn't the first day of a month, a
simple subtraction and followed by a type cast that I don't grasp
would do the trick.
===================================================================== Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx
What I'm looking for is guidance on date arithmetic.
cknell@xxxxxxxxxx - email
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