GROUP BY with MAX(DATE) [duplicate]

Answer 1

SELECT train, dest, time FROM ( 
  SELECT train, dest, time, 
    RANK() OVER (PARTITION BY train ORDER BY time DESC) dest_rank
    FROM traintable
  ) where dest_rank = 1

Answer 2

Here's an example that only uses a Left join and I believe is more efficient than any group by method out there: ExchangeCore Blog

SELECT t1.*
FROM TrainTable t1 LEFT JOIN TrainTable t2
ON (t1.Train = t2.Train AND t1.Time < t2.Time)
WHERE t2.Time IS NULL;

Answer 3

Another solution:

select * from traintable
where (train, time) in (select train, max(time) from traintable group by train);

Answer 4

You cannot include non-aggregated columns in your result set which are not grouped. If a train has only one destination, then just add the destination column to your group by clause, otherwise you need to rethink your query.

Try:

SELECT t.Train, t.Dest, r.MaxTime
FROM (
      SELECT Train, MAX(Time) as MaxTime
      FROM TrainTable
      GROUP BY Train
) r
INNER JOIN TrainTable t
ON t.Train = r.Train AND t.Time = r.MaxTime

Answer 5

As long as there are no duplicates (and trains tend to only arrive at one station at a time)...

select Train, MAX(Time),
      max(Dest) keep (DENSE_RANK LAST ORDER BY Time) max_keep
from TrainTable
GROUP BY Train;

Answer 6

I know I'm late to the party, but try this...

SELECT 
    `Train`, 
    `Dest`,
    SUBSTRING_INDEX(GROUP_CONCAT(`Time` ORDER BY `Time` DESC), ",", 1) AS `Time`
FROM TrainTable
GROUP BY Train;

Src: Group Concat Documentation

Edit: fixed sql syntax