Subject: Re: Iteration of for() for first 100 intergers.
From: Hermann Stamm-Wilbrandt <STAMMW@xxxxxxxxxx>
Date: Thu, 18 Nov 2010 21:42:13 +0100
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> 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
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From: Michael Kay <mike@xxxxxxxxxxxx>
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Date: 11/18/2010 05:22 PM
Subject: Re: 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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