[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: XSLT 2.0/XPath 2.0 Date arithmetic
> 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/
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|