I have a database table in the following format.
Product | Date | Score
A | 01/01/18 | 99
B | 01/01/18 | 98
C | 01/01/18 | 97
--------------------------
A | 02/01/18 | 99
B | 02/01/18 | 98
C | 02/01/18 | 97
--------------------------
D | 03/01/18 | 99
A | 03/01/18 | 98
B | 03/01/18 | 97
C | 03/01/18 | 96
I want to pick the first from every month such that there are no repeat products. For example, the output of the above table should be
Product | Date | Score
A | 01/01/18 | 99
B | 02/01/18 | 98
D | 03/01/18 | 99
How do I get this result with a single sql query? The actual table is much bigger than this and I want top 20 from every month without repetition.
This is a hard problem -- a type of subgraph problem that isn't really suitable to SQL. There is a brute force approach:
with jan as (
select *
from t
where date = '2018-01-01'
limit 1
),
feb as (
select *
from t
where date = '2018-02-01' and
product not in (select product from jan)
),
mar as (
select *
from t
where date = '2018-03-01' and
product not in (select product from jan) and
product not in (select product from feb)
)
select *
from jan
union all
select *
from feb
union all
select *
from mar;
You can generalize this with additional CTEs. But there is no guarantee that a month will have a product -- even when it could have had one.
It is possible by using row_number.
select * from (
select row_Number() over(partition by Product order by Product ) as rno,* from
Products
) as t where t.rno<=20
I think you want top 20 records every month without repeating products than below solution will be work.
select *
into #temp
from
(values
('A','01/01/18','99')
,('B','01/01/18','98')
,('C','01/01/18','97')
,('A','02/01/18','99')
,('B','02/01/18','98')
,('C','02/01/18','97')
,('D','03/01/18','99')
,('A','03/01/18','98')
,('B','03/01/18','97')
,('C','03/01/18','96')
) AS VTE (Product ,Date, Score )
select * from
(
select * , ROW_NUMBER() over (partition by date,product order by score ) as rn
from #TEMP
)
A where rn < 20
+-------+-------+-----------+
| EmpID | PerID | VisitDate |
+-------+-------+-----------+
| 1 | 22 | 2/24/2017 |
| 1 | 22 | 3/25/2017 |
| 1 | 22 | 4/5/2017 |
| 2 | 33 | 5/6/2017 |
| 2 | 33 | 8/9/2017 |
| 2 | 33 | 6/7/2017 |
+-------+-------+-----------+
I am trying to find the latest visit date and average days between visits per EmpID. For Avg, I'll first have to order the days consecutively and then find the average.
Eg: Avg. days for EmpID=1 and PerID=22 would be [29(Days between 3/25 and 2/24) + 11 (Days between 3/25 and 4/5)/2] = 20 Days.
Desired Output:
+-------+-------+----------+----------+
| EmpID | PerID | MaxVDate | AvgVDays |
+-------+-------+----------+----------+
| 1 | 22 | 4/5/2017 | 20 |
| 2 | 33 | 8/9/2017 | 47.5 |
+-------+-------+----------+----------+
Attempt:
SELECT
EmpID
,PerID
,MAX(VisitDate) AS MaxVDate
,--Dunno how to find average AS AvgVDays
FROM
T1
GROUP BY
EmpID
,PerID
You can use lag to get the previous date and compute the date difference. Then use avg window function to get the average days.
Select distinct empid,perid,maxVdate,avg(diff_with_prev) OVER(Partition by empid) as avgVDays
from (
SELECT EmpID,PerID
,MAX(VisitDate) OVER(Partition BY EmpID) AS MaxVDate
,DATEDIFF(DAY,LAG(VisitDate) OVER(Partition BY EmpID order by VisitDate), VisitDate) as diff_with_prev
FROM T1
) t
Here's an option...
IF OBJECT_ID('tempdb..#TestData', 'U') IS NOT NULL
DROP TABLE #TestData;
CREATE TABLE #TestData (
EmpID INT NOT NULL,
PerID INT NOT NULL,
VisitDate DATE NOT NULL
);
INSERT #TestData (EmpID, PerID, VisitDate) VALUES
(1, 22, '2/24/2017'),
(1, 22, '3/25/2017'),
(1, 22, '4/5/2017'),
(2, 33, '5/6/2017'),
(2, 33, '8/9/2017'),
(2, 33, '6/7/2017');
-- SELECT * FROM #TestData td;
SELECT
db.EmpID,
db.PerID,
AvgDays = AVG(db.DaysBetween * 1.0)
FROM (
SELECT
*,
DaysBetween = DATEDIFF(dd, LAG(td.VisitDate, 1) OVER (PARTITION BY td.EmpID, td.PerID ORDER BY td.VisitDate), td.VisitDate)
FROM
#TestData td
) db
GROUP BY
db.EmpID,
db.PerID;
Results...
EmpID PerID AvgDays
----------- ----------- ---------------------------------------
1 22 20.000000
2 33 47.500000
The task is much easier than you think. You get the average with (last visit - first visit) / (count visits - 1).
select
empid,
perid,
max(VisitDate) as MaxVDate,
datediff(day, min(VisitDate), max(VisitDate)) * 1.0 / (count(*) - 1) as avgvdays
from mytable
group by empid, perid
having count(*) > 1
order by empid, perid;
The multiplication with 1.0 is necessary in order to avoid integer division. (You could also cast to decimal instead.)
As the calcualtion only makes sense for empid/perid pairs with more than one entry (and in order to avoid division by zero), I have applied an according HAVING clause.
Here is a test: http://rextester.com/AIFPA62612
I have a table like this one:
Yr | Mnth | W_ID | X_ID | Y_ID | Z_ID | Purchases | Sales | Returns |
2015 | 10 | 1 | 5210 | 1402 | 2 | 1000.00 | etc | etc |
2015 | 12 | 1 | 5210 | 1402 | 2 | 12000.00 | etc | etc |
2016 | 1 | 1 | 5210 | 1402 | 2 | 1000.00 | etc | etc |
2016 | 3 | 1 | 5210 | 1402 | 2 | etc | etc | etc |
2014 | 3 | 9 | 880 | 2 | 7 | etc | etc | etc |
2014 | 12 | 9 | 880 | 2 | 7 | etc | etc | etc |
2015 | 5 | 9 | 880 | 2 | 7 | etc | etc | etc |
2015 | 7 | 9 | 880 | 2 | 7 | etc | etc | etc |
For each combination of (W, X, Y, Z) I would like to insert the months that don't appear in the table and are between the first and last month.
In this example, for combination (W=1, X=5210, Y=1402, Z=2), I would like to have additional rows for 2015/11 and 2016/02, where Purchases, Sales and Returns are NULL. For combination (W=9, X=880, Y=2, Z=7) I would like to have additional rows for months between 2014/4 and 2014/11, 2015/01 and 2015/04, 2016/06.
I hope I have explained myself correctly.
Thank you in advance for any help you can provide.
The process is rather cumbersome in this case, but quite possible. One method uses a recursive CTE. Another uses a numbers table. I'm going to use the latter.
The idea is:
Find the minimum and maximum values for the year/month combination for each set of ids. For this, the values will be turned into months since time 0 using the formula year*12 + month.
Generate a bunch of numbers.
Generate all rows between the two values for each combination of ids.
For each generated row, use arithmetic to re-extract the year and month.
Use left join to bring in the original data.
The query looks like:
with n as (
select row_number() over (order by (select null)) - 1 as n -- start at 0
from master.spt_values
),
minmax as (
select w_id, x_id, y_id, z_id, min(yr*12 + mnth) as minyyyymm,
max(yr*12 + mnth) as maxyyyymm
from t
group by w_id, x_id, y_id, z_id
),
wxyz as (
select minmax.*, minmax.minyyyymm + n.n,
(minmax.minyyyymm + n.n) / 12 as yyyy,
((minmax.minyyyymm + n.n) % 12) + 1 as mm
from minmax join
n
on minmax.minyyyymm + n.n <= minmax.maxyyyymm
)
select wxyz.yyyy, wxyz.mm, wxyz.w_id, wxyz.x_id, wxyz.y_id, wxyz.z_id,
<columns from t here>
from wxyz left join
t
on wxyz.w_id = t.w_id and wxyz.x_id = t.x_id and wxyz.y_id = t.y_id and
wxyz.z_id = t.z_id and wxyz.yyyy = t.yr and wxyz.mm = t.mnth;
Thank you for your help.
Your solution works, but I noticed it is not very good in terms of performance, but meanwhile I have managed to get a solution for my problem.
DECLARE #start_date DATE, #end_date DATE;
SET #start_date = (SELECT MIN(EOMONTH(DATEFROMPARTS(Yr , Mnth, 1))) FROM Table_Input);
SET #end_date = (SELECT MAX(EOMONTH(DATEFROMPARTS(Yr , Mnth, 1))) FROM Table_Input);
DECLARE #tdates TABLE (Period DATE, Yr INT, Mnth INT);
WHILE #start_date <= #end_date
BEGIN
INSERT INTO #tdates(PEriod, Yr, Mnth) VALUES(#start_date, YEAR(#start_date), MONTH(#start_date));
SET #start_date = EOMONTH(DATEADD(mm,1,DATEFROMPARTS(YEAR(#start_date), MONTH(#start_date), 1)));
END
DECLARE #pks TABLE (W_ID NVARCHAR(50), X_ID NVARCHAR(50)
, Y_ID NVARCHAR(50), Z_ID NVARCHAR(50)
, PerMin DATE, PerMax DATE);
INSERT INTO #pks (W_ID, X_ID, Y_ID, Z_ID, PerMin, PerMax)
SELECT W_ID, X_ID, Y_ID, Z_ID
, MIN(EOMONTH(DATEFROMPARTS(Ano, Mes, 1))) AS PerMin
, MAX(EOMONTH(DATEFROMPARTS(Ano, Mes, 1))) AS PerMax
FROM Table1
GROUP BY W_ID, X_ID, Y_ID, Z_ID;
INSERT INTO Table_Output(W_ID, X_ID, Y_ID, Z_ID
, ComprasLiquidas, RTV, DevManuais, ComprasBrutas, Vendas, Stock, ReceitasComerciais)
SELECT TP.DB, TP.Ano, TP.Mes, TP.Supplier_Code, TP.Depart_Code, TP.BizUnit_Code
, TA.ComprasLiquidas, TA.RTV, TA.DevManuais, TA.ComprasBrutas, TA.Vendas, TA.Stock, TA.ReceitasComerciais
FROM
(
SELECT W_ID, X_ID, Y_ID, Z_ID
FROM #tdatas CROSS JOIN #pks
WHERE Period BETWEEN PerMin And PerMax
) AS TP
LEFT JOIN Table_Input AS TA
ON TP.W_ID = TA.W_ID AND TP.X_ID = TA.X_ID AND TP.Y_ID = TA.Y_ID
AND TP.Z_ID = TA.Z_ID
AND TP.Yr = TA.Yr
AND TP.Mnth = TA.Mnth
ORDER BY TP.W_ID, TP.X_ID, TP.Y_ID, TP.Z_ID, TP.Yr, TP.Mnth;
I do the following:
Get the Min and Max date of the entire table - #start_date and #end_date variables;
Create an auxiliary table with all dates between Min and Max - #tdates table;
Get all the combinations of (W_ID, X_ID, Y_ID, Z_ID) along with the min and max dates of that combination - #pks table;
Create the cartesian product between #tdates and #pks, and in the WHERE clause I filter the results between the Min and Max of the combination;
Compute a LEFT JOIN of the cartesian product table with the input data table.
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)