Most recent date with maximum score - mysql

I have the table definition below.
CREATE TABLE IF NOT EXISTS ranking (
user_id int(11) unsigned NOT NULL,
create_date date NOT NULL,
score double(8,2),
PRIMARY KEY (user_id, create_date)
)
insert into ranking (user_id, create_date, score) values
(1, '2017-03-01', 100),
(1, '2017-03-02', 90),
(1, '2017-03-03', 80),
(1, '2017-03-04', 100),
(1, '2017-03-05', 90),
(2, '2017-03-01', 90),
(2, '2017-03-02', 80),
(2, '2017-03-03', 100),
(2, '2017-03-5', 100),
(3, '2017-03-01', 80),
(3, '2017-03-02', 100),
(3, '2017-03-03', 90),
(3, '2017-03-6', 100);
select * from ranking;
user_id | create_date | score
1 | 2017-03-01 | 100
1 | 2017-03-02 | 90
1 | 2017-03-03 | 80
1 | 2017-03-04 | 100
1 | 2017-03-05 | 90
2 | 2017-03-01 | 90
2 | 2017-03-02 | 80
2 | 2017-03-03 | 100
2 | 2017-03-05 | 100
3 | 2017-03-01 | 80
3 | 2017-03-02 | 100
3 | 2017-03-03 | 90
3 | 2017-03-06 | 100
What I want is for each user_id, get the most recent create_date on which the score is maximum. For example, in the example above, for user_id = 1, when create_date = 2017-03-01 and create_date = 2017-03-04, the maximum score is 100, but I just want the most recent date with the maximum score, i.e., create_date = 2017-03-04 and score = 100. The query result is as follows:
user_id | create_date | score
1 | 2017-03-04 | 100
2 | 2017-03-05 | 100
3 | 2017-03-06 | 100
Below is my solution, which returns the expected result but I believe there exist better solutions.
SELECT a.* from
(
SELECT s1.user_id , s1.create_date, s1.score FROM ranking AS s1
INNER JOIN
(SELECT user_id , FORMAT(max(score), 0) as best_score FROM ranking GROUP BY user_id ) AS s2
ON s1.user_id = s2.user_id AND s1.score = s2.best_score
) a
NATURAL LEFT JOIN
(
SELECT s1.user_id , s1.create_date, s1.score FROM ranking AS s1
INNER JOIN
(
SELECT user_id , create_date, score FROM ranking
) s2
WHERE s1.user_id = s2.user_id AND s1.score = s2.score AND s1.create_date < s2.create_date
) b
WHERE b.user_id IS NULL;
Can someone provide better solutions? Thanks.


SELECT t1.user_id,
MAX(t1.create_date) AS max_date,
t2.max_score
FROM ranking t1
INNER JOIN
(
SELECT user_id, MAX(score) AS max_score
FROM ranking
GROUP BY user_id
) t2
ON t1.user_id = t2.user_id AND
t1.score = t2.max_score
GROUP BY t1.user_id
Output:
Demo here:
Rextester


Try this:
select user_id, max(create_date),max(score) from ranking GROUP BY user_id
result:
1 2017-03-04 100.00
2 2017-03-05 100.00
3 2017-03-06 100.00
or
select user_id, max(create_date),cast(max(score) as UNSIGNED) as maxscore from ranking GROUP BY user_id
result:
1 2017-03-04 100
2 2017-03-05 100
3 2017-03-06 100


Try this query -
SELECT r1.* FROM ranking r1
JOIN (SELECT user_id, MAX(score) max_score FROM ranking GROUP BY user_id) r2
ON r1.user_id = r2.user_id AND r1.score = r2.max_score
JOIN (SELECT user_id, score, MAX(create_date) max_create_date FROM ranking GROUP BY user_id, score) r3
ON r1.user_id = r3.user_id AND r1.score = r3.score AND r1.create_date = r3.max_create_date;
1 04-Mar-17 100
2 05-Mar-17 100
3 06-Mar-17 100

Related

MySQL Select entire row from matching Group By value

Lets say I have 2 tables like so:
MyTable1:
Name ID Timestamp TestNum Grade
Alex 1101 2020-10-01 12:00:00 1 85
Alex 1101 2020-10-02 13:00:00 2 90
Alex 1101 2020-10-03 8:00:00 3 95
Alex 1101 2020-10-04 10:00:00 4 90
MyTable2:
ID Avg StDev
1101 90 4.08
I am trying to get the row of the first (Timestamp) instance where the grade was X standard deviations away.
ExpectedResults:
Name ID Timestamp TestNum StDevsAway
Alex 1101 2020-10-01 12:00:00 1 -1.23
Alex 1101 2020-10-02 13:00:00 2 0
Alex 1101 2020-10-03 8:00:00 3 1.23
The 4th row should not be returned as its Standard Deviations Away was already found at a previous Timestamp.
I'm still fairly new to MySQL, but this is where I'm at so far:
select a.Name
, a.ID
, a.Timestamp
, a.TestNum
, round( ( a.Grade - b.Avg ) / b.StDev, 2 ) as StDevsAway
from MyTable1 as a
join MyTable2 as b
on a.ID = b.ID
group
by round( ( a.Grade - b.Avg ) / b.StDev, 2 );

I think the question is just about finding "first" row for each id/grade tupe. So (asssuming MySQL 8.0):
select t1.*
from (
select t1.*, row_number() over(partition by id, grade order by timestamp) rn
from mytable1 t1
) t1
where rn = 1
Then, you can bring the second table with a join if you like:
select t1.*, round(t1.grade - t2.avg) / t2.stdev, 2) stdevsaway
from (
select t1.*, row_number() over(partition by id, grade order by timestamp) rn
from mytable1 t1
) t1
inner join mytable2 t2 on t2.id = t1.id
where rn = 1
In earlier versions, you can filter with a subquery:
select t1.*, round(t1.grade - t2.avg) / t2.stdev, 2) stdevsaway
from mytable1 t1
inner join mytable2 t2 on t2.id = t1.id
where t1.timestamp = (
select min(t11.timestamp) from mytable1 t11 where t11.id = t1.id and t11.grade = t1.grade
)

In previous Versin and of Course in mysql8 as well you can do this.
this will exclude every testnum, that gas standard deviation,except the first one, for that user
Schema (MySQL v5.5)
CREATE TABLE MyTable1 (
`Name` VARCHAR(4),
`ID` INTEGER,
`Timestamp` DATETIME,
`TestNum` VARCHAR(7),
`Grade` INTEGER
);
INSERT INTO MyTable1
(`Name`, `ID`, `Timestamp`, `TestNum`, `Grade`)
VALUES
('Alex', '1101', '2020-10-01 12:00:00', '1', '85'),
('Alex', '1101', '2020-10-02 13:00:00', '2', '90'),
('Alex', '1101', '2020-10-03 08:00:00', '3','95'),
('Alex', '1101', '2020-10-04 10:00:00', '4', '90');
CREATE TABLE MyTable2 (
`ID` INTEGER,
`Avg` INTEGER,
`StDev` FLOAT
);
INSERT INTO MyTable2
(`ID`, `Avg`, `StDev`)
VALUES
('1101', '90', '4.08');
Query #1
select
a.Name
, a.ID
, a.Timestamp
, a.TestNum
, round( ( a.Grade - b.Avg ) / b.StDev, 2 ) as StDevsAway
from MyTable1 as a join MyTable2 as b on a.ID = b.ID
WHERE
TestNum NOT IN (SELECT TestNum
FROM MyTable1 c
WHERE c.`ID` = a.`ID`
AND c.`Grade` = b.Avg
AND c.`TestNum`<> (SELECT MIN(TestNum)
FROM MyTable1 d
WHERE d.`ID` = a.`ID`
AND d.`Grade` = b.Avg)
);
| Name | ID | Timestamp | TestNum | StDevsAway |
| ---- | ---- | ------------------- | ------- | ---------- |
| Alex | 1101 | 2020-10-01 12:00:00 | 1 | -1.23 |
| Alex | 1101 | 2020-10-02 13:00:00 | 2 | 0 |
| Alex | 1101 | 2020-10-03 08:00:00 | 3 | 1.23 |
View on DB Fiddle

Solving max Scoring player from each group in mySQL

Refer to another Stack Overflow question here, however the answers there didn't include the group_id 3 player.
I tried to replicate the answer in MySQL but I am not familiar with PostgreSQL. Anyone can show how to proceed it in MySQL?
The question is to return the max scored player as winner_id from each group
create table players (
player_id integer not null unique,
group_id integer not null
);
create table matches (
match_id integer not null unique,
first_player integer not null,
second_player integer not null,
first_score integer not null,
second_score integer not null
);
insert into players values(20, 2);
insert into players values(30, 1);
insert into players values(40, 3);
insert into players values(45, 1);
insert into players values(50, 2);
insert into players values(65, 1);
insert into matches values(1, 30, 45, 10, 12);
insert into matches values(2, 20, 50, 5, 5);
insert into matches values(13, 65, 45, 10, 10);
insert into matches values(5, 30, 65, 3, 15);
insert into matches values(42, 45, 65, 8, 4);
matches table
match_id | first_player | second_player | first_score | second_score
----------+--------------+---------------+-------------+--------------
1 | 30 | 45 | 10 | 12
2 | 20 | 50 | 5 | 5
13 | 65 | 45 | 10 | 10
5 | 30 | 65 | 3 | 15
42 | 45 | 65 | 8 | 4
Expected output
group_id | winner_id
----------+-----------
1 | 45
2 | 20
3 | 40

I presume that since you can't use the solution to the other question that you are using MySQL 5.7 or below. In that case, you have to simulate the ROW_NUMBER/PARTITION functionality, which you can do with a LEFT JOIN from a derived table of scores per player with itself, joining on the score being greater than that in the first table. Any player who has no scores greater in the joined table clearly has the highest score. Since there can be ties, we then take the minimum of the player_id values from that table (when there is no tie, this has no effect).
SELECT group_id, MIN(player_id) AS player_id
FROM (
SELECT t1.group_id, t1.player_id
FROM (
SELECT p.player_id, p.group_id,
SUM(CASE WHEN m.first_player = p.player_id THEN m.first_score
ELSE m.second_score
END) AS score
FROM players p
LEFT JOIN matches m ON m.first_player = p.player_id OR m.second_player = p.player_id
GROUP BY p.player_id, p.group_id
) t1
LEFT JOIN (
SELECT p.player_id, p.group_id,
SUM(CASE WHEN m.first_player = p.player_id THEN m.first_score
ELSE m.second_score
END) AS score
FROM players p
LEFT JOIN matches m ON m.first_player = p.player_id OR m.second_player = p.player_id
GROUP BY p.player_id, p.group_id
) t2 ON t2.group_id = t1.group_id AND t2.score > t1.score
GROUP BY t1.group_id, t1.player_id
HAVING COUNT(t2.player_id) = 0
) w
GROUP BY group_id
Output:
group_id player_id
1 45
2 20
3 40
Demo on db-fiddle

SQL MAX sensor values based on timestamp and multiple group by [duplicate]

I have a table of player performance:
CREATE TABLE TopTen (
id INT UNSIGNED PRIMARY KEY AUTO_INCREMENT,
home INT UNSIGNED NOT NULL,
`datetime`DATETIME NOT NULL,
player VARCHAR(6) NOT NULL,
resource INT NOT NULL
);
What query will return the rows for each distinct home holding its maximum value of datetime? In other words, how can I filter by the maximum datetime (grouped by home) and still include other non-grouped, non-aggregate columns (such as player) in the result?
For this sample data:
INSERT INTO TopTen
(id, home, `datetime`, player, resource)
VALUES
(1, 10, '04/03/2009', 'john', 399),
(2, 11, '04/03/2009', 'juliet', 244),
(5, 12, '04/03/2009', 'borat', 555),
(3, 10, '03/03/2009', 'john', 300),
(4, 11, '03/03/2009', 'juliet', 200),
(6, 12, '03/03/2009', 'borat', 500),
(7, 13, '24/12/2008', 'borat', 600),
(8, 13, '01/01/2009', 'borat', 700)
;
the result should be:
id
home
datetime
player
resource
1
10
04/03/2009
john
399
2
11
04/03/2009
juliet
244
5
12
04/03/2009
borat
555
8
13
01/01/2009
borat
700
I tried a subquery getting the maximum datetime for each home:
-- 1 ..by the MySQL manual:
SELECT DISTINCT
home,
id,
datetime AS dt,
player,
resource
FROM TopTen t1
WHERE `datetime` = (SELECT
MAX(t2.datetime)
FROM TopTen t2
GROUP BY home)
GROUP BY `datetime`
ORDER BY `datetime` DESC
The result-set has 130 rows although database holds 187, indicating the result includes some duplicates of home.
Then I tried joining to a subquery that gets the maximum datetime for each row id:
-- 2 ..join
SELECT
s1.id,
s1.home,
s1.datetime,
s1.player,
s1.resource
FROM TopTen s1
JOIN (SELECT
id,
MAX(`datetime`) AS dt
FROM TopTen
GROUP BY id) AS s2
ON s1.id = s2.id
ORDER BY `datetime`
Nope. Gives all the records.
I tried various exotic queries, each with various results, but nothing that got me any closer to solving this problem.

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

The fastest MySQL solution, without inner queries and without GROUP BY:
SELECT m.* -- get the row that contains the max value
FROM topten m -- "m" from "max"
LEFT JOIN topten b -- "b" from "bigger"
ON m.home = b.home -- match "max" row with "bigger" row by `home`
AND m.datetime < b.datetime -- want "bigger" than "max"
WHERE b.datetime IS NULL -- keep only if there is no bigger than max
Explanation:
Join the table with itself using the home column. The use of LEFT JOIN ensures all the rows from table m appear in the result set. Those that don't have a match in table b will have NULLs for the columns of b.
The other condition on the JOIN asks to match only the rows from b that have bigger value on the datetime column than the row from m.
Using the data posted in the question, the LEFT JOIN will produce this pairs:
+------------------------------------------+--------------------------------+
| the row from `m` | the matching row from `b` |
|------------------------------------------|--------------------------------|
| id home datetime player resource | id home datetime ... |
|----|-----|------------|--------|---------|------|------|------------|-----|
| 1 | 10 | 04/03/2009 | john | 399 | NULL | NULL | NULL | ... | *
| 2 | 11 | 04/03/2009 | juliet | 244 | NULL | NULL | NULL | ... | *
| 5 | 12 | 04/03/2009 | borat | 555 | NULL | NULL | NULL | ... | *
| 3 | 10 | 03/03/2009 | john | 300 | 1 | 10 | 04/03/2009 | ... |
| 4 | 11 | 03/03/2009 | juliet | 200 | 2 | 11 | 04/03/2009 | ... |
| 6 | 12 | 03/03/2009 | borat | 500 | 5 | 12 | 04/03/2009 | ... |
| 7 | 13 | 24/12/2008 | borat | 600 | 8 | 13 | 01/01/2009 | ... |
| 8 | 13 | 01/01/2009 | borat | 700 | NULL | NULL | NULL | ... | *
+------------------------------------------+--------------------------------+
Finally, the WHERE clause keeps only the pairs that have NULLs in the columns of b (they are marked with * in the table above); this means, due to the second condition from the JOIN clause, the row selected from m has the biggest value in column datetime.
Read the SQL Antipatterns: Avoiding the Pitfalls of Database Programming book for other SQL tips.

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.

Since people seem to keep running into this thread (comment date ranges from 1.5 year) isn't this much simpler:
SELECT * FROM (SELECT * FROM topten ORDER BY datetime DESC) tmp GROUP BY home
No aggregation functions needed...
Cheers.

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

Try this for SQL Server:
WITH cte AS (
SELECT home, MAX(year) AS year FROM Table1 GROUP BY home
)
SELECT * FROM Table1 a INNER JOIN cte ON a.home = cte.home AND a.year = cte.year

Here is MySQL version which prints only one entry where there are duplicates MAX(datetime) in a group.
You could test here http://www.sqlfiddle.com/#!2/0a4ae/1
Sample Data
mysql> SELECT * from topten;
+------+------+---------------------+--------+----------+
| id | home | datetime | player | resource |
+------+------+---------------------+--------+----------+
| 1 | 10 | 2009-04-03 00:00:00 | john | 399 |
| 2 | 11 | 2009-04-03 00:00:00 | juliet | 244 |
| 3 | 10 | 2009-03-03 00:00:00 | john | 300 |
| 4 | 11 | 2009-03-03 00:00:00 | juliet | 200 |
| 5 | 12 | 2009-04-03 00:00:00 | borat | 555 |
| 6 | 12 | 2009-03-03 00:00:00 | borat | 500 |
| 7 | 13 | 2008-12-24 00:00:00 | borat | 600 |
| 8 | 13 | 2009-01-01 00:00:00 | borat | 700 |
| 9 | 10 | 2009-04-03 00:00:00 | borat | 700 |
| 10 | 11 | 2009-04-03 00:00:00 | borat | 700 |
| 12 | 12 | 2009-04-03 00:00:00 | borat | 700 |
+------+------+---------------------+--------+----------+
MySQL Version with User variable
SELECT *
FROM (
SELECT ord.*,
IF (#prev_home = ord.home, 0, 1) AS is_first_appear,
#prev_home := ord.home
FROM (
SELECT t1.id, t1.home, t1.player, t1.resource
FROM topten t1
INNER JOIN (
SELECT home, MAX(datetime) AS mx_dt
FROM topten
GROUP BY home
) x ON t1.home = x.home AND t1.datetime = x.mx_dt
ORDER BY home
) ord, (SELECT #prev_home := 0, #seq := 0) init
) y
WHERE is_first_appear = 1;
+------+------+--------+----------+-----------------+------------------------+
| id | home | player | resource | is_first_appear | #prev_home := ord.home |
+------+------+--------+----------+-----------------+------------------------+
| 9 | 10 | borat | 700 | 1 | 10 |
| 10 | 11 | borat | 700 | 1 | 11 |
| 12 | 12 | borat | 700 | 1 | 12 |
| 8 | 13 | borat | 700 | 1 | 13 |
+------+------+--------+----------+-----------------+------------------------+
4 rows in set (0.00 sec)
Accepted Answers' outout
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
+------+------+---------------------+--------+----------+
| id | home | datetime | player | resource |
+------+------+---------------------+--------+----------+
| 1 | 10 | 2009-04-03 00:00:00 | john | 399 |
| 2 | 11 | 2009-04-03 00:00:00 | juliet | 244 |
| 5 | 12 | 2009-04-03 00:00:00 | borat | 555 |
| 8 | 13 | 2009-01-01 00:00:00 | borat | 700 |
| 9 | 10 | 2009-04-03 00:00:00 | borat | 700 |
| 10 | 11 | 2009-04-03 00:00:00 | borat | 700 |
| 12 | 12 | 2009-04-03 00:00:00 | borat | 700 |
+------+------+---------------------+--------+----------+
7 rows in set (0.00 sec)

SELECT c1, c2, c3, c4, c5 FROM table1 WHERE c3 = (select max(c3) from table)
SELECT * FROM table1 WHERE c3 = (select max(c3) from table1)

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

Why not using:
SELECT home, MAX(datetime) AS MaxDateTime,player,resource FROM topten GROUP BY home
Did I miss something?

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:
id
home
date
player
resource
1
10
2009-03-04 00:00:00
john
399
2
11
2009-03-04 00:00:00
juliet
244
5
12
2009-03-04 00:00:00
borat
555
8
13
2009-01-01 00:00:00
borat
700
db<>fiddle here

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)
In the ORDER BY clause, it uses a window function to generate & sort by a ROW_NUMBER - assigning a 1 value to the highest [datetime] for each [home].
SELECT TOP 1 WITH TIES will then select one record with the lowest ROW_NUMBER (which will be 1), as well as all records with a tying ROW_NUMBER (also 1)
As a consequence, you retrieve all data for each of the 1st ranked records - that is, all data for records with the highest [datetime] value with their given [home] value.

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

#Michae The accepted answer will working fine in most of the cases but it fail for one for as below.
In case if there were 2 rows having HomeID and Datetime same the query will return both rows, not distinct HomeID as required, for that add Distinct in query as below.
SELECT DISTINCT tt.home , tt.MaxDateTime
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

this is the query you need:
SELECT b.id, a.home,b.[datetime],b.player,a.resource FROM
(SELECT home,MAX(resource) AS resource FROM tbl_1 GROUP BY home) AS a
LEFT JOIN
(SELECT id,home,[datetime],player,resource FROM tbl_1) AS b
ON a.resource = b.resource WHERE a.home =b.home;

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)

How to get biggest differences in column compared to last month?

I have two tables - Products and Prices
Every month table Prices is populated with new prices for each products. How can I get 5 products whose prices have the biggest incremental difference from last month prices?
table Products
id | name
1 | apples
2 | pears
3 | bananas
table Prices
id | price | product_id | created_at
1 | 10 | 1 | 2017-02-07 07:00:00
2 | 10 | 2 | 2017-02-07 07:00:00
3 | 15 | 3 | 2017-02-07 07:00:00
5 | 15 | 1 | 2017-03-07 07:00:00
6 | 20 | 2 | 2017-03-07 07:00:00
7 | 25 | 3 | 2017-03-07 07:00:00
The result would be to find out that
1. Bananas has prices by 15 higher (lastMonth: 15, now: 25)
2. Pears 2 has prices by 10 higher (lastMonth: 10, now: 20)
3. Apples has prices by 5 higher (lastMonth: 10, now: 15)
I was thinking something like this (uff I know this is terrible)
SELECT products.id, products.name, prices.beforePrice, prices.afterPrice, prices.difference
FROM products
INNER JOIN prices ON products.id = prices.product_id
WHERE
(
SELECT *biggest-difference*
FROM prices
WHERE *difference_between_last_2_months*
GROUP BY product_id
LIMIT 5
)

Create table/insert data
CREATE TABLE Products
(`id` INT, `name` VARCHAR(7))
;
INSERT INTO Products
(`id`, `name`)
VALUES
(1, 'apples'),
(2, 'pears'),
(3, 'bananas')
;
CREATE TABLE Prices
(`id` INT, `price` INT, `product_id` INT, `created_at` DATETIME)
;
INSERT INTO Prices
(`id`, `price`, `product_id`, `created_at`)
VALUES
(1, 10, 1, '2017-02-07 07:00:00'),
(2, 10, 2, '2017-02-07 07:00:00'),
(3, 15, 3, '2017-02-07 07:00:00'),
(5, 15, 1, '2017-03-07 07:00:00'),
(6, 20, 2, '2017-03-07 07:00:00'),
(7, 25, 3, '2017-03-07 07:00:00')
;
Query
SELECT
Products.id
, Products.name
, (current_month.price - last_month.price) AS difference
, (
CASE
WHEN last_month.price > current_month.price
THEN 'lower'
WHEN last_month.price < current_month.price
THEN 'higher'
END
) AS incremental
, last_month.price 'lastMonth'
, current_month.price 'now'
FROM (
SELECT
*
FROM
Prices
WHERE
MONTH(created_at) = MONTH((CURDATE() - INTERVAL 1 MONTH))
)
AS
last_month
INNER JOIN (
SELECT
*
FROM
Prices
WHERE
MONTH(created_at) = MONTH((CURDATE()))
)
AS
current_month
ON
last_month.product_id = current_month.product_id
INNER JOIN
Products
ON
last_month.product_id = Products.id
WHERE
last_month.price < current_month.price #incremental should be higher
ORDER BY
difference DESC
LIMIT 5
Result
id name difference incremental lastMonth now
------ ------- ---------- ----------- --------- --------
2 pears 10 higher 10 20
3 bananas 10 higher 15 25
1 apples 5 higher 10 15

You can use a proper joins based on fliterd select by month.
This should return the value you need ( you can add the literal string you need )
select p.name, m1.price as this_month, m2.price as prev_month, m2.price-m1.price as diff
from product
left join (
select price, product_id
from Prices
where month(created_at) = month(NOW())
and year(created_at) = year(NOW())
) m1 on m1.product_id = p.id
left join (
select price, product_id
from Prices
where month(created_at) = MONT(DATE_SUB(NOW(), INTERVAL 1 MONTH)
and year(created_at) = year(DATE_SUB(NOW(), INTERVAL 1 MONTH)
) m2 on m2.product_id = p.id
order by diff desc
limit 5

SQL - get latest records from table where field is unique

I have a table of data as follows
id status conversation_id message_id date_created
1 1 1 72 2012-01-01 00:00:00
2 2 1 87 2012-03-03 00:00:00
3 2 2 95 2012-05-05 00:00:00
I want to get all the rows from the table in date_created DESC order, but only one row per conversation_id. So in the case of the example data above, I would want to get the rows with id 2 and 3.
Any advice is much appreciated.

SELECT t.id, t.status, t.conversation_id, t.message_id, t.date_created
FROM YourTable t
INNER JOIN (SELECT conversation_id, MAX(date_created) AS MaxDate
FROM YourTable
GROUP BY conversation_id) q
ON t.conversation_id = q.conversation_id
AND t.date_created = q.MaxDate
ORDER BY t.date_created DESC;

See SQL Fiddle
SELECT T.*
FROM T
WHERE NOT EXISTS (
SELECT *
FROM T AS _T
WHERE _T.conversation_id = T.conversation_id
AND (
_T.date_created > T.date_created
OR
_T.date_created = T.date_created AND _T.id > T.id)
)
ORDER BY T.date_created DESC
gets
ID STATUS CONVERSATION_ID MESSAGE_ID DATE_CREATED
3 2 2 95 May, 05 2012
2 2 1 87 March, 03 2012

Creation and inserts borrowed from #Incidently:
CREATE TABLE T
(id int, status int, conversation_id int, message_id int, date_created datetime);
insert into T (id, status, conversation_id, message_id, date_created)
values(1, 1, 1, 72, '2012-01-01 00:00:00');
insert into T (id, status, conversation_id, message_id, date_created)
values(2, 2, 1, 87, '2012-03-03 00:00:00');
insert into T (id, status, conversation_id, message_id, date_created)
values(3, 2 , 2 , 95 , '2012-05-05 00:00:00');
Supposing the id is the primary key then this solves the problem pointed by #Incidentally in #Joe's answer:
select T.*
from
T
inner join (
select conversation_id, max(id) as id
from T
group by conversation_id
) s on s.id = T.id
order by T.date_created desc, T.conversation_id
;
+------+--------+-----------------+------------+---------------------+
| id | status | conversation_id | message_id | date_created |
+------+--------+-----------------+------------+---------------------+
| 3 | 2 | 2 | 95 | 2012-05-05 00:00:00 |
| 2 | 2 | 1 | 87 | 2012-03-03 00:00:00 |
+------+--------+-----------------+------------+---------------------+