Is there a [straightforward] way to order results *first*, *then* group by another column, with SQL?

Question

I see that in SQL, the GROUP BY has to precede ORDER BY expression. Does this imply that ordering is done after grouping discards identical rows/columns?

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Answer 1

create table user_payments
(
    phone_nr int NOT NULL,
    payed_until_ts datetime NOT NULL
)

insert into user_payments
(phone_nr, payed_until_ts)
values
(1, '2016-01-28'), -- today
(1, '2016-01-27'), -- yesterday  
(2, '2016-01-27'), -- yesterday 
(2, '2016-01-29')  -- tomorrow

select phone_nr, MAX(payed_until_ts) as latest_payment
from user_payments
group by phone_nr

-- OUTPUT:
-- phone_nr latest_payment
-- 1        2016-01-28 00:00:00.000
-- 2        2016-01-29 00:00:00.000

In the above example, I have used datetime column but similar query should work for timestamp column.

The MAX function will basically do the "ORDER BY" payed_until_ts column and pick the latest value for each phone_nr. Also, you will get only one value for each phone_nr due to "GROUP BY" clause.