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[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 "$today" isn't the first day of a month, a simple subtraction and followed by a type cast that I don't grasp would do the trick. If you're using Xalan and Java, you can always use the (non-standard) extensions. Something along the following lines (not tested): <?xml version="1.0" ?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:java="http://xml.apache.org/xalan/java" exclude-result-prefixes="java"> ... <xsl:variable name="today">20060517</xsl:variable> <xsl:variable name="sdf" select="java:java.text.SimpleDateFormat.new('yyyymmdd')" /> <xsl:variable name="tMinus1" select="java:format($sdf, (1000*60*60*24)+java:Long(java:parse($sdf, $today)))" /> ... </xsl:stylesheet> See http://xml.apache.org/xalan-j/extensions.html#ext-func-calls JavaScript is also available/possible 
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