find(array("memberships" => array('$in' => new MongoId($user_id))));
This will not only return the membership and related membership for possible
Whereas with plan 1 it is two queries.
On 6 October 2011 14:42, Karl Seguin <karlseg...@gmail.com> wrote:
> I think you are over simplifying.--
> How well does #2 work in a sharded environment if you want to get all of a
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