I need to compare current and next values of the node var2 [and var3]
and find the first occurrence when var2[position()] = var2[next, i.e.
position()+1]
Are you saying that is impossible to do in XSLT ?
It's almost certainly possible to do what you want but your decription
is pretty hard to follow. You need to describe the transformation that
you want to do not just use bits of XPath and hope that we understand
what you want. Xpath's position() function for example does not relate
to the position of a node within a tree (a node can have any number of
values returned buy position(), depending on how the node is
selected). I'm _guessing_ that what you want to know is the sibling order
of the node in the input tree in which case you might want
following-sibling::*[1] but that is just a guess.
OK, lets try to get a position of the first occurrence when var2 = max node.
As I said, I don't want to change the order of the nodes in a tree with sort.
What would be a correct syntax ?
<xsl:for-each select=mystruct/myarray1[1]/myvar>
<xsl:if var2=*[last()]>
<xsl:value-of select="position()">
</xsl:if>
</xsl:for-each>
The goal is to trim the table [see my previous post],
get rid of multiple identical occurrences of rebate1 & rebate2.
Do you understand me ?
Thank you,
Oleg.
I want from inside for-each loop mystruct/myarray1[i]/myvar
[or does it require recursion?]
compare the values is the parallel branch,
of parallel "current" node ../../myarray2[i]/myvar/var3 and its "next node".
Again, where:
mystruct/myarray1[i]/myvar/var2 and
mystruct/myarray2[i]/myvar/var3
Again you are just typing in syntactically correct XPath and asking "can
I do that" since the Xpath you typed above is syntactically correct then
clearly you can do that, but I have no idea if that will do what you
want, as you've not said what you are trying to do. For example do your
elements with name myarray1 have child elements with name i ? If they do
not, then the above XPath's will select nothing.
Is there an easy way to find out the max value of the node (child elements)
without sorting it ?
The sequence of nodes I am dealing with is supposed to be in increasing
order,
If you know it's already sorted, just take the last
select="*[last()]"
or in your case take the last of each then you only have two itemsto
compare which you can do with > to work out the overall maximum.
David
________________________________________________________________________
This e-mail has been scanned for all viruses by Star. The
service is powered by MessageLabs. For more information on a proactive
anti-virus service working around the clock, around the globe, visit:
http://www.star.net.uk
________________________________________________________________________
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail:
<mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--
--~------------------------------------------------------------------
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--