Subject: Re: [xsl] Iteration of for() for first 100 intergers. From: Hermann Stamm-Wilbrandt <STAMMW@xxxxxxxxxx> Date: Thu, 18 Nov 2010 21:42:13 +0100 |
> In XPath 1.0 > > you're out of luck. There's a workaround to iterate over a node-set > containing n nodes, and use position(). See 1b in this document for enhancements of that technique by Wendel Piez: http://www.xml.org//sites/www.xml.org/files/xslt_efficient_programming_techniques.pdf Mit besten Gruessen / Best wishes, Hermann Stamm-Wilbrandt Developer, XML Compiler, L3 Fixpack team lead WebSphere DataPower SOA Appliances ---------------------------------------------------------------------- IBM Deutschland Research & Development GmbH Vorsitzender des Aufsichtsrats: Martin Jetter Geschaeftsfuehrung: Dirk Wittkopp Sitz der Gesellschaft: Boeblingen Registergericht: Amtsgericht Stuttgart, HRB 243294 From: Michael Kay <mike@xxxxxxxxxxxx> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Date: 11/18/2010 05:22 PM Subject: Re: [xsl] Iteration of for() for first 100 intergers. On 18/11/2010 16:09, ram wrote: > Hi, > I am looking for simple for() which will iterate through first 100 numbers. how can i do that. > > for(int i=0;i<=100;i++){ > System.out.println("each value"+i); > } > In XPath 2.0: for $i in 1 to 100 return ..... In XPath 1.0 you're out of luck. There's a workaround to iterate over a node-set containing n nodes, and use position(). Michael Kay Saxonica
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