Subject: RE: XSLT 2.0/XPath 2.0 Date arithmetic
From: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Thu, 18 May 2006 08:27:54 +0100
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> 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 "$today"
> isn't the first day of a month, a simple subtraction and
> followed by a type cast that I don't grasp would do the trick.
>
> What I'm looking for is guidance on date arithmetic.
You'd be much better off working with the xs:date type, which uses the
format 2006-05-17.
So, two functions to convert between your non-standard dates and standard
xs:date objects:
<xsl:import-schema>
<xs:schema target-namespace="http://my-date">
<xs:simpleType name="yyyymmdd-date">
<xs:restriction base="xs:string">
<xs:pattern value="[0-9]{4}[0-1][0-9][0-3][0-9]"/>
</
</
</
</
<xsl:function name="f:to-iso-date" as="xs:date">
<xsl:param name="in" as="my:yyyymmdd-date"/>
<xsl:sequence select="xs:date(replace($in, '(\d{4})(\d{2})(\d{2})',
'$1-$2-$3'))"/>
</xsl:function>
<xsl:function name="f:to-yyyymmdd-date" as="my:yyyymmdd-date">
<xsl:param name="in" as="xs:date"/>
<xsl:sequence select="my:yyyymmdd-date(translate(string($in, '-', ''))"/>
</xsl:function>
then:
select="f:to-yyyymmdd-date(f:to-iso-date($input-date) -
xs:dayTimeDuration('PT1D'))"
If you're not schema-aware, then use xs:string in place of my:yyyymmdd-date
- all you lose is type-checking.
Michael Kay
http://www.saxonica.com/
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