On Jun 17, 2014, at 7:38 AM, Walter Lee Davis <[email protected]> wrote:
> Your code is correct: some_reference.order('kind DESC') is the correct way to
> get the order you are asking for. Remember that if you only have three
> possible values, the implicit sort of the members within each kind will
> probably be the ID, all other things being equal.
There is no "implicit" sort in SQL, the order within each kind will be
indeterminate.
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Scott Ribe
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