How can I SELECT rows with MAX(Column value), PARTITION by another column in MYSQL?

You are so close! All you need to do is select BOTH the home and its max date time, then join back to the topten table on BOTH fields:

SELECT tt.*
FROM topten tt
INNER JOIN
    (SELECT home, MAX(datetime) AS MaxDateTime
    FROM topten
    GROUP BY home) groupedtt 
ON tt.home = groupedtt.home 
AND tt.datetime = groupedtt.MaxDateTime

Here goes T-SQL version:

-- Test data
DECLARE @TestTable TABLE (id INT, home INT, date DATETIME, 
  player VARCHAR(20), resource INT)
INSERT INTO @TestTable
SELECT 1, 10, '2009-03-04', 'john', 399 UNION
SELECT 2, 11, '2009-03-04', 'juliet', 244 UNION
SELECT 5, 12, '2009-03-04', 'borat', 555 UNION
SELECT 3, 10, '2009-03-03', 'john', 300 UNION
SELECT 4, 11, '2009-03-03', 'juliet', 200 UNION
SELECT 6, 12, '2009-03-03', 'borat', 500 UNION
SELECT 7, 13, '2008-12-24', 'borat', 600 UNION
SELECT 8, 13, '2009-01-01', 'borat', 700

-- Answer
SELECT id, home, date, player, resource 
FROM (SELECT id, home, date, player, resource, 
    RANK() OVER (PARTITION BY home ORDER BY date DESC) N
    FROM @TestTable
)M WHERE N = 1

-- and if you really want only home with max date
SELECT T.id, T.home, T.date, T.player, T.resource 
    FROM @TestTable T
INNER JOIN 
(   SELECT TI.id, TI.home, TI.date, 
        RANK() OVER (PARTITION BY TI.home ORDER BY TI.date) N
    FROM @TestTable TI
    WHERE TI.date IN (SELECT MAX(TM.date) FROM @TestTable TM)
)TJ ON TJ.N = 1 AND T.id = TJ.id

EDIT
Unfortunately, there are no RANK() OVER function in MySQL.
But it can be emulated, see Emulating Analytic (AKA Ranking) Functions with MySQL.
So this is MySQL version:

SELECT id, home, date, player, resource 
FROM TestTable AS t1 
WHERE 
    (SELECT COUNT(*) 
            FROM TestTable AS t2 
            WHERE t2.home = t1.home AND t2.date > t1.date
    ) = 0

This will work even if you have two or more rows for each home with equal DATETIME's:

SELECT id, home, datetime, player, resource
FROM   (
       SELECT (
              SELECT  id
              FROM    topten ti
              WHERE   ti.home = t1.home
              ORDER BY
                      ti.datetime DESC
              LIMIT 1
              ) lid
       FROM   (
              SELECT  DISTINCT home
              FROM    topten
              ) t1
       ) ro, topten t2
WHERE  t2.id = ro.lid

I think this will give you the desired result:

SELECT   home, MAX(datetime)
FROM     my_table
GROUP BY home

BUT if you need other columns as well, just make a join with the original table (check Michael La Voie answer)

Best regards.

You can also try this one and for large tables query performance will be better. It works when there no more than two records for each home and their dates are different. Better general MySQL query is one from Michael La Voie above.

SELECT t1.id, t1.home, t1.date, t1.player, t1.resource
FROM   t_scores_1 t1 
INNER JOIN t_scores_1 t2
   ON t1.home = t2.home
WHERE t1.date > t2.date

Or in case of Postgres or those dbs that provide analytic functions try

SELECT t.* FROM 
(SELECT t1.id, t1.home, t1.date, t1.player, t1.resource
  , row_number() over (partition by t1.home order by t1.date desc) rw
 FROM   topten t1 
 INNER JOIN topten t2
   ON t1.home = t2.home
 WHERE t1.date > t2.date 
) t
WHERE t.rw = 1

SELECT  tt.*
FROM    TestTable tt 
INNER JOIN 
        (
        SELECT  coord, MAX(datetime) AS MaxDateTime 
        FROM    rapsa 
        GROUP BY
                krd 
        ) groupedtt
ON      tt.coord = groupedtt.coord
        AND tt.datetime = groupedtt.MaxDateTime

This works on Oracle:

with table_max as(
  select id
       , home
       , datetime
       , player
       , resource
       , max(home) over (partition by home) maxhome
    from table  
)
select id
     , home
     , datetime
     , player
     , resource
  from table_max
 where home = maxhome

Another way to gt the most recent row per group using a sub query which basically calculates a rank for each row per group and then filter out your most recent rows as with rank = 1

select a.*
from topten a
where (
  select count(*)
  from topten b
  where a.home = b.home
  and a.`datetime` < b.`datetime`
) +1 = 1

DEMO

Here is the visual demo for rank no for each row for better understanding

By reading some comments what about if there are two rows which have same 'home' and 'datetime' field values?

Above query will fail and will return more than 1 rows for above situation. To cover up this situation there will be a need of another criteria/parameter/column to decide which row should be taken which falls in above situation. By viewing sample data set i assume there is a primary key column id which should be set to auto increment. So we can use this column to pick the most recent row by tweaking same query with the help of CASE statement like

select a.*
from topten a
where (
  select count(*)
  from topten b
  where a.home = b.home
  and  case 
       when a.`datetime` = b.`datetime`
       then a.id < b.id
       else a.`datetime` < b.`datetime`
       end
) + 1 = 1

DEMO

Above query will pick the row with highest id among the same datetime values

visual demo for rank no for each row

In MySQL 8.0 this can be achieved efficiently by using row_number() window function with common table expression.

(Here row_number() basically generating unique sequence for each row for every player starting with 1 in descending order of resource. So, for every player row with sequence number 1 will be with highest resource value. Now all we need to do is selecting row with sequence number 1 for each player. It can be done by writing an outer query around this query. But we used common table expression instead since it's more readable.)

Schema:

 create  TABLE TestTable(id INT, home INT, date DATETIME, 
   player VARCHAR(20), resource INT);
 INSERT INTO TestTable
 SELECT 1, 10, '2009-03-04', 'john', 399 UNION
 SELECT 2, 11, '2009-03-04', 'juliet', 244 UNION
 SELECT 5, 12, '2009-03-04', 'borat', 555 UNION
 SELECT 3, 10, '2009-03-03', 'john', 300 UNION
 SELECT 4, 11, '2009-03-03', 'juliet', 200 UNION
 SELECT 6, 12, '2009-03-03', 'borat', 500 UNION
 SELECT 7, 13, '2008-12-24', 'borat', 600 UNION
 SELECT 8, 13, '2009-01-01', 'borat', 700

Query:

 with cte as 
 (
     select id, home, date , player, resource, 
     Row_Number()Over(Partition by home order by date desc) rownumber from TestTable
 )
 select id, home, date , player, resource from cte where rownumber=1

Output:

db<>fiddle here

Try this

select * from mytable a join
(select home, max(datetime) datetime
from mytable
group by home) b
 on a.home = b.home and a.datetime = b.datetime

Regards K

Hope below query will give the desired output:

Select id, home,datetime,player,resource, row_number() over (Partition by home ORDER by datetime desc) as rownum from tablename where rownum=1

(NOTE: The answer of Michael is perfect for a situation where the target column datetime cannot have duplicate values for each distinct home.)

If your table has duplicate rows for homexdatetime and you need to only select one row for each distinct home column, here is my solution to it:

Your table needs one unique column (like id). If it doesn't, create a view and add a random column to it.

Use this query to select a single row for each unique home value. Selects the lowest id in case of duplicate datetime.

SELECT tt.*
FROM topten tt
INNER JOIN
    (
    SELECT min(id) as min_id, home from topten tt2
    INNER JOIN 
        (
        SELECT home, MAX(datetime) AS MaxDateTime
        FROM topten
        GROUP BY home) groupedtt2
    ON tt2.home = groupedtt2.home
    ) as groupedtt
ON tt.id = groupedtt.id

Accepted answer doesn't work for me if there are 2 records with same date and home. It will return 2 records after join. While I need to select any (random) of them. This query is used as joined subquery so just limit 1 is not possible there. Here is how I reached desired result. Don't know about performance however.

select SUBSTRING_INDEX(GROUP_CONCAT(id order by datetime desc separator ','),',',1) as id, home, MAX(datetime) as 'datetime'
 from topten
 group by (home)

Because this hasn't been posted - this works in SQLServer, and is the only solution I've seen that doesn't require subqueries or CTEs - I think this is the most elegant way to solve this kind of problem

  SELECT TOP 1 WITH TIES *
    FROM TopTen
ORDER BY ROW_NUMBER() OVER (PARTITION BY home
                                ORDER BY [datetime] DESC)

Some notes on how it works - The Window Function in the Order By clause applies a counter to each group of home values, such that the one with the highest [datetime] value receives 1.

By SELECTing TOP 1 WITH TIES , you're selecting the record with the first ROW_NUMBER value (which is 1), as well as all other records with the same 'tying' ROW_NUMBER value of 1.

As a consequence, you retrieve all data for each of the 1st ranked records.