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(_at_)saxonica(_dot_)com>
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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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