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Re: [xsl] more elegant way of doing this? (very simple)

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unless you really need to always use a node test 
AContribution rather than name()='AContribution'
It's namespace aware and likely more efficient.

Are you using xsl 1 or 2?

xpath 1:

sum(Records/Record/*[self::AContribution|self::BContribution|self::CContribution][number()=number()])

xpath2 you can do the same, or a bit more simply:

sum(Records/Record/(AContribution|BContribution|CContribution)[number()=number()])

David

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