Andy,



At 02:49 PM 4/2/2008, you wrote:
> From: David Carlisle [mailto:davidc@xxxxxxxxx]
> Sent: 02 April, 2008 14:37
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: Re: [xsl] Splitting files 5 at a time
>
>
>
>    [count() mod 5 = 0]
>
> [position() mod 5 = 1]

I thought I understood his problem to be that he wanted the
first five customers in one file, then the next five customers
in the next, etc.  If he just did what you suggested wouldn't
he just grab every 1 in 5 customers out of the group of five
customers?

Customer  1 => position() mod 5 = 1
Customer  2 => position() mod 5 = 2
Customer  3 => position() mod 5 = 3
Customer  4 => position() mod 5 = 4
Customer  5 => position() mod 5 = 0

Customer  6 => position() mod 5 = 1
Customer  7 => position() mod 5 = 2
Customer  8 => position() mod 5 = 3
Customer  9 => position() mod 5 = 4
Customer 10 => position() mod 5 = 0

So if I understood your solution, he would create the first file
with Customer 1, 6, 11, 16, 21 and not Customer 1, 2, 3, 4, 5.
Is that correct?

Not quite, although you are correct as to which nodes would be 
selected by David's XPath.



A more complete explanation would offer that the for-each selects nos 
1, 6, 11 etc. and that a file should be generated for each of these, 
to contain the customer selected plus its four immediately following 
customer siblings.



So the OP would need to select customer[position() mod 5 = 1] to 
group the customers into their files, and then include the results of 
processing (.|following-sibling[position() < 5]) into each file generated.



David was just enjoying shorthand, as he frequently does, and perhaps 
trusting that the OP would fill in the rest.



Cheers,
Wendell


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
Mulberry Technologies, Inc.                http://www.mulberrytech.com
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