Re: [xsl] Iteration of for() for first 100 intergers.

[Hermann Stamm-Wilbrandt <STAMMW@xxxxxxxxxx>]

> 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: [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