MySQL query help (semi-unique results) - sql

user_id | user_destination | user_date_out | user_date_in | user_purpose | uid
0095 | NYC | 2010-11-25 | 2010-11-26 | Work | 1
0105 | Seattle | 2010-11-15 | 2010-11-20 | Work | 2
0095 | Home | 2010-11-10 | 2010-11-11 | Personal | 3
0123 | Nashville | 2010-11-10 | 2010-11-12 | Doctober | 4
I have the above data in a MySQL table. I need a query that will output the row if it is the most recent user_date_out for that user_id. Hard to explain, but the rows that should be displayed via the query are uid 2,3 and 4 (UID 1 would fall off because the user_date_out is "older" than UID 3 for the same user). Can anyone help me with this? Thanks in advance!
EDIT: I solved it by using the following query:
SELECT *
FROM `table`
WHERE `uid` = (
SELECT `uid`
FROM `table` as `alt`
WHERE `alt`.`user_id` = `table`.`user_id`
ORDER BY `user_date_out`
LIMIT 1
)
ORDER BY `user_date_out`


Per request
SELECT *
FROM `table`
WHERE `uid` = (
SELECT `uid`
FROM `table` as `alt`
WHERE `alt`.`user_id` = `table`.`user_id`
ORDER BY `user_date_out`
LIMIT 1
)
ORDER BY `user_date_out`


SELECT t.*, a_subquery.min_user_date_out
FROM your_table t
JOIN (SELECT user_id, MIN(user_date_out) AS min_user_date_out
FROM your_table
GROUP BY user_id) a_subquery
ON a_subquery.min_user_date_out = t.user_date_out
AND a_subquery.user_id = t.user_id
Since MySQL sometimes has issues w/ subqueries, here's the same equivalent query with a self-exclusion join:
SELECT t.*
FROM your_table t
LEFT JOIN your_table t2
ON t.user_id = t2.user_id
AND t.user_date_out > t2.user_date_out
WHERE t2.user_id IS NULL
Note that both of these queries will not work if you have more than one of any user_date_out for a given user_id.

Related

Tie-breaking mutliple matches on MAX() in SQL

I have a table that looks like this:
| client_id | program_id | provider_id | date_of_service | data_entry_date | data_entry_time |
| --------- | ---------- | ----------- | --------------- | --------------- | --------------- |
| 2 | 5 | 6 | 02/02/2022 | 02/02/2022 | 0945 |
| 2 | 5 | 6 | 02/02/2022 | 02/07/2022 | 0900 |
| 2 | 5 | 6 | 02/04/2022 | 02/04/2022 | 1000 |
| 2 | 5 | 6 | 02/04/2022 | 02/04/2022 | 1700 |
| 2 | 5 | 6 | 02/04/2022 | 02/05/2022 | 0800 |
| 2 | 5 | 6 | 02/04/2022 | 02/05/2022 | 0900 |
I need to get the most recent date_of_service entered. From the table above, the desired result/row is:
date_of_service = 02/04/2022, data_entry_date = 02/05/2022, data_entry_time = 0900
This resulting date_of_service will be left joined to the master table.
This query mostly works:
SELECT t1.client_id, t1.program_id, t1.provider_id, t2.date_of_service
FROM table1 as t1
WHERE provider_id = '6'
LEFT JOIN
(SELECT client_id, program_id, provider_id, date_of_service
FROM table2) as t2
ON t2.client_id = t1.client_id
AND t2.program_id = t1.program_id
AND t2.provider_id = t1.provider_id
AND t2.date_of_service =
(SELECT MAX(date_of_service)
FROM t2 as t3
WHERE t3.client_id = t1.client_id
AND t3.program_id = t1.program_id
AND t3.provider_id = t1.provider_id
)
)
But it also returns multiple rows whenever there is more than one match on the max(date_of_service).
To solve this, I need to use the max data_entry_date to break any ties whenever there is more than one row that matches the max(date_of_service). Likewise, I also need to use the max data_entry_time to break any ties whenever there is more than one row that also matches the max data_entry_date.
I tried the following:
SELECT t1.client_id, t1.program_id, t1.provider_id, t2.date_of_service
FROM table1 as t1
WHERE provider_id = '6'
LEFT JOIN
(SELECT TOP(1) client_id, program_id, provider_id, date_of_service, data_entry_date, data_entry_time
FROM table2
ORDER BY date_of_service DESC, data_entry_date DESC, data_entry_time DESC
) as t2
ON t2.client_id = t1.client_id
AND t2.program_id = t1.program_id
AND t2.provider_id = t1.provider_id
But I can only get it to return null values for the date_of_service.
Likewise, this:
SELECT t1.client_id, t1.program_id, t1.provider_id, t2.date_of_service
FROM table1 as t1
WHERE provider_id = '6'
LEFT JOIN
(
SELECT TOP(1) client_id AS client_id2, program_id AS program_id2, provider_id AS provider_id2, date_of_service, data_entry_date, data_entry_time
FROM table2 AS t3
JOIN
(SELECT
MAX(date_of_service) AS max_date_of_service
,MAX(data_entry_date) AS max_data_entry_date
FROM table1
WHERE date_of_service = (SELECT MAX(date_of_service) FROM table2)
) AS t4
ON t3.date_of_service = t4.max_date_of_service
AND t3.data_entry_date = t4.max_data_entry_date
ORDER BY data_entry_time
) AS t2
ON t2.client_id2 = t1.client_id
AND t2.program_id2 = t1.program_id
AND t2.provider_id2 = t1.provider_id
... works (meaning it doesn't throw any errors), but it only seems to return null values for me.
I've tried various combinations of MAX, ORDER BY, and multiple variations of JOIN's, but haven't found one that works yet.
I don't know what version my SQL database is, but it doesn't appear to handle window functions like OVER and PARTITION or other things like COALESCE. I've been using DBeaver 22.2.0 to test the SQL scripts.

Based on your what you've provided, looks like you can simply query table2:
SELECT client_id, program_id, provider_id, MAX(date_of_service), MAX(data_entry_date), MAX(data_entry_time)
FROM table2
GROUP BY client_id, program_id, provider_id
If you need to join this result set to table1, just JOIN to the statement above on client_id, program_id, provider_id

Try using below query. This is using just joins and sub query.
SELECT TOP 1 * FROM table1 t1
JOIN (
SELECT
MAX(date_of_Service) AS Max_date_of_Service
,MAX(data_entry_date) AS Max_data_entry_date
FROM table1
WHERE date_of_Service = (SELECT MAX(date_of_Service) FROM table1)
)t2
ON t1.date_of_Service = t2.Max_date_of_Service
AND t1.data_entry_date = t2.Max_data_entry_date
ORDER BY data_entry_time

Finding created on dates for duplicates in SQL

I have one table of contact records and I'm trying to get the count of duplicate records that were created on each date. I'm not looking to include the original instance in the count. I'm using SQL Server.
Here's an example table
| email | created_on |
| ------------- | ---------- |
| aaa#email.com | 08-16-22 |
| bbb#email.com | 08-16-22 |
| zzz#email.com | 08-16-22 |
| bbb#email.com | 07-12-22 |
| aaa#email.com | 07-12-22 |
| zzz#email.com | 06-08-22 |
| aaa#email.com | 06-08-22 |
| bbb#email.com | 04-21-22 |
And I'm expecting to return
| created_on | dupe_count |
| ---------- | ---------- |
| 08-16-22 | 3 |
| 07-12-22 | 2 |
| 06-08-22 | 0 |
| 04-21-22 | 0 |
Edited to add error message:
error message

I created a sub table based on email and created date row number. Then, you query that, and ignore the date when the email first was created (row number 1). Works perfectly fine in this case.
Entire code:
Create table #Temp
(
email varchar(50),
dateCreated date
)
insert into #Temp
(email, dateCreated) values
('aaa#email.com', '08-16-22'),
('bbb#email.com', '08-16-22'),
('zzz#email.com', '08-16-22'),
('bbb#email.com', '07-12-22'),
('aaa#email.com', '07-12-22'),
('zzz#email.com', '06-08-22'),
('aaa#email.com', '06-08-22'),
('bbb#email.com', '04-21-22')
select datecreated, sum(case when r = 1 then 0 else 1 end) as duplicates
from
(
Select email, datecreated, ROW_NUMBER() over(partition by email
order by datecreated) as r from #Temp
) b
group by dateCreated
drop table #Temp
Output:
datecreated duplicates
2022-04-21 0
2022-06-08 0
2022-07-12 2
2022-08-16 3

You can calculate the difference between total count of emails for every day and the count of unique emails for the day:
select created_on,
count(email) - count(distinct email) as dupe_count
from cte
group by created_on
It seems I have misunderstood your request, and you wanted to consider previous created_on dates' too:
ct as (
select created_on,
(select case when (select count(*)
from cte t2
where t1.email = t2.email and t1.created_on > t2.created_on
) > 0 then email end) as c
from cte t1)
select created_on,
count(distinct c) as dupe_count
from ct
group by created_on
order by 1
It seems that in oracle it is also possible to aggregate it using one query:
select created_on,
count(distinct case when (select count(*)
from cte t2
where t1.email = t2.email and t1.created_on > t2.created_on
) > 0 then email end) as c
from cte t1
group by created_on
order by 1

postgresql - How to get one row the min value

I have table (t_image) with this column
datacd | imagecode | indexdate
----------------------------------
A | 1 | 20170213
A | 2 | 20170213
A | 3 | 20170214
B | 4 | 20170201
B | 5 | 20170202
desired result is this
datacd | imagecode | indexdate
----------------------------------
A | 1 | 20170213
B | 4 | 20170201
In the above table, I want to retrieve 1 row for each datacd who has the minimum index date
Here is my query, but the result returns 2 rows for datacd A
select *
from (
select datacd, min(indexdate) as indexdate
from t_image
group by datacd
) as t1 inner join t_image as t2 on t2.datacd = t1.datacd and t2.indexdate = t1.indexdate;

The Postgres proprietary distinct on () operator is typically the fastest solution for greatest-n-per-group queries:
select distinct on (datacd) *
from t_image
order by datacd, indexdate;

One option uses ROW_NUMBER():
SELECT t.datacd,
t.imagecode,
t.indexdate
FROM
(
SELECT datacd, imagecode, indexdate,
ROW_NUMBER() OVER (PARTITION BY datacd ORDER BY indexdate) rn
FROM t_image
) t
WHERE t.rn = 1

A very basic SQL issue I'm stuck with [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 query range?

Raw Data
| ID | STATUS |
| 1 | A |
| 2 | A |
| 3 | B |
| 4 | B |
| 5 | B |
| 6 | A |
| 7 | A |
| 8 | A |
| 9 | C |
Result
| START | END |
| 1 | 2 |
| 6 | 8 |
Range of STATUS A
How to query ?

This should give you the correct ranges:
SELECT
STATUS,
MIN(ID),
max_id
FROM (
SELECT
t1.STATUS,
t1.ID,
COALESCE(MAX(t2.ID), t1.ID) max_id
FROM
yourtable t1 LEFT JOIN yourtable t2
ON t1.STATUS=t2.STATUS AND t1.ID<t2.ID
WHERE
NOT EXISTS (SELECT NULL
FROM yourtable t3
WHERE
t3.STATUS!=t1.STATUS
AND t3.ID>t1.ID AND t3.ID<t2.ID)
GROUP BY
t1.ID,
t1.STATUS
) s
WHERE
status = 'A'
GROUP BY
STATUS,
max_id
Please see fiddle here.

You are probably better off with a cursor-based solution or a client-side function.
However, if you were using Oracle - the following would work.
WITH LOWER_VALS AS
( -- All the Ids with no immediate predecessor
SELECT ROWNUM AS RN, STATUS, ID AS LOWER FROM
(
SELECT STATUS, ID
FROM RAWDATA RD1
WHERE RD1.ID -1 NOT IN
(SELECT ID FROM RAWDATA PRED_TABLE WHERE PRED_TABLE.STATUS = RD1.STATUS)
ORDER BY STATUS, ID
)
) ,
UPPER_VALS AS
( -- All the Ids with no immediate successor
SELECT ROWNUM AS RN, STATUS, ID AS UPPER FROM
(
SELECT STATUS, ID
FROM RAWDATA RD2
WHERE RD2.ID +1 NOT IN
(SELECT ID FROM RAWDATA SUCC_TABLE WHERE SUCC_TABLE.STATUS = RD2.STATUS)
ORDER BY STATUS, ID
)
)
SELECT
L.STATUS, L.LOWER, U.UPPER
FROM
LOWER_VALS L
JOIN UPPER_VALS U ON
U.RN = L.RN;
Results in the set
A 1 2
A 6 8
B 3 5
C 9 9
http://sqlfiddle.com/#!4/10184/2

There is not a lot to go on from what you put, but I think this might work. I am using T-SQL because I don't know what you are using?
SELECT
min(ID)
, max(ID)
FROM RawData
WHERE [Status] = 'A'