[xquery-talk] return only latest version of an item
A. Steven Anderson
steve at asanderson.com
Thu May 14 19:17:08 PDT 2009
>
> Which must be a bug. saxon returns a different result.
>
Perhaps, or I've lost something in translation from my simplified example.
> Using distinct-values to group by id could be another option, though maybe
> not the fastest:
>
> for $id in distinct-values(/items//id)
> let $itemsById := /items/item[id = $id]
> return
> $itemsById[version = max($itemsById/version)]
>
> You should have an index on id and version in any case.
>
This worked! It's twice as fast as my original convoluted solution.
Many thanks to all who responded!!
--
A. Steven Anderson
Independent Consultant
steve at asanderson.com
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