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Re: [xsl] HST's answers Re: [xsl] Efficient way to check sequence membership -

2011-03-03 04:50:32

It seems the following expression can do checks based on node identity
(I inferred this meaning from Dave's requirement... the "member()"
function),

some $x in $seq satisfies ($x is $item)



DP's requirement statement could be read that way, but he also said the aim was to solve HT's problem, which doesn't involve nodes or node identity.

But for completeness, if you want to do this for node identity, you can use

test="$item intersect $seq"

Michael Kay
Saxonica

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