RE: Newbie question: Incrementing the position in the

Sorry, I'm short for time, but another thought:

If you have the same amount of cells for every row, you could just do a
position() mod numberOfCells = 0 triggers a closing and opening tr element.

for-each
if test="position() mod 4 = 0
	<tr/>
process cells normally
end for-each

I hope I have understood your dilemma correctly.



-----Original Message-----
From: Kenny Akridge [mailto:kenny@xxxxxxxxxxxxxxxxx] 
Sent: Monday, April 12, 2004 3:48 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE:  Newbie question: Incrementing the position in the tree
midstream

It sounds like you might want to do some recursion and pass in N recursively
to the template.  Then you could access cell[n], cell[n+1]... 

Just one idea.  There are a lot of approaches to this one.

-----Original Message-----
From: Durston, Andrew (AGRE) [mailto:andrew.durston@xxxxxxxxxxxxxxxxxxxx] 
Sent: Monday, April 12, 2004 3:42 PM
To: Xsl-List (E-mail)
Subject:  Newbie question: Incrementing the position in the tree
midstream


Hi all,

Just getting started with XSL and all and had a question about whether you
can change positions in the tree midstream.
I have a set of XML outputs from a database which splits a table into
individual cells -

test-plan/object/cell N
test-plan/object/cell N+1
test-plan/object/cell N+2
test-plan/object/cell N+3

I can search to find cell N. I'd like to be able to print cell N, cell N+1,
cell N+2 ... all at once and then when XSL goes back
to searching (via a For) through the tree, it skips N+1, N+2 etc. (basically
does a table row and then skips to the beginning of
the next row). Without entering another XML attribute into our DB to
indicate beginning and end of rows... How do I increment
the position, moving from object/cell N to object/cell N+1 midstream?

Thanks in advance,
Andrew

Andrew C. Durston
Smiths Aerospace
Electronic Systems - Grand Rapids
616.241.8715
andrew.durston@xxxxxxxxxxxxxxxxxxxx



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