SQL：如何选择每天的最大记录？

Question

I found a lot of similar questions but no one fits perfectly for my case and I am struggling for hours for a solution. My table is composed by the fields DAY, HOUR, EVENT1, EVENT2, EVENT3. Therefore I have 24 rows each day. EVENT1, EVENT2, EVENT3 have some values and I'd like to select each day only the row (I mean the record) for which EVENT3 has the maximum value in the day (among the 24 hours). The final outcome will be one row per day

Answer 1

One method uses a correlated subquery:

select t.*
from t
where t.event3 = (select max(t2.event3)
                  from t t2
                  where t2.date = t.date
                 );

In most databases, this has very good performance with an index on (date, event3) .

A more canonical solution uses row_number() :

select t.*
from (select t.*,
             row_number() over (partition by date order by event3 desc) as seqnum
      from t
     ) t
where seqnum = 1;

Answer 2

Another option aside from using correlated subqueries is to write this is a left self-join, something like this:

SELECT t.*
FROM t
LEFT JOIN t AS t2 ON t.day = t2.day AND t2.event3 > t.event3
WHERE t2.id IS NULL

If you want to select an arbitrary matching row each day in the event of multiple rows with the same maximum event3, tack GROUP BY t.day on the end of that.

I'm not sure how performance of this is going to compare to Gordon Linoff's solutions, but they might get assembled into quite similar query plans by the RDBMS anyway.