I am working with a table in Databricks Delta lake. It gets new records appended every month. The field insert_dt indicates when the records are inserted.
| ID | Mrc | insert_dt |
|----|-----|------------|
| 1 | 40 | 2022-01-01 |
| 2 | 30 | 2022-01-01 |
| 3 | 50 | 2022-01-01 |
| 4 | 20 | 2022-02-01 |
| 5 | 45 | 2022-02-01 |
| 6 | 55 | 2022-03-01 |
Now I want to aggregate by insert_dt and calculate the average of Mrc. For each date, the average is done not just for the records of that date but all records with date prior to that. In this example, there are 3 rows for 2022-01-01, 5 rows for 2022-02-01 and 6 rows for 2022-03-01. The expected results would look like this:
| Mrc | insert_dt |
|-----|------------|
| 40 | 2022-01-01 |
| 37 | 2022-02-01 |
| 40 | 2022-03-01 |
How do I write a query to do that?
I checked the documentation for Delta-lake databricks (https://docs.databricks.com/sql/language-manual/sql-ref-window-functions.html ) and it looks like TSQL so I think this will work for you, but you may need to tweak slightly.
The approach is to condense each day to a single point and then use window functions to get the running totals. Note that any given day may have a different count, so you can't just average the averages.
--Enter the sample data you gave as a CTE for testing
;with cteSample as (
SELECT * FROM ( VALUES
(1, 40, CONVERT(date,'2022-01-01'))
, ('2', '30', '2022-01-01')
, ('3', '50', '2022-01-01')
, ('4', '20', '2022-02-01')
, ('5', '45', '2022-02-01')
, ('6', '55', '2022-03-01')
) as TabA(ID, Mrc, insert_dt)
)--Solution begins here, find the total and count for each date
--because window can only handle a single "last row"
, cteGrouped as (
SELECT insert_dt, SUM(Mrc) as MRCSum, COUNT(*) as MRCCount
FROM cteSample
GROUP BY insert_dt
)--Now use the window function to get the totals "up to today"
, cteTotals as (
SELECT insert_dt
, SUM(MRCSum) OVER (ORDER BY insert_dt RANGE
BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS MrcSum
, SUM(MRCCount) OVER (ORDER BY insert_dt RANGE
BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS MrcCount
FROM cteGrouped as G
) --Now divide out to get the average to date
SELECT insert_dt, MrcSum/MrcCount as MRCAverage
FROM cteTotals as T
This gives the following output
insert_dt
MRCAverage
2022-01-01
40
2022-02-01
37
2022-03-01
40
Calculate a running average using a window function (the inner subquery) and then pick only one row per insert_dt - the one with the highest id. I only tested this on PostgreSQL 13 so not sure how far does delta-lake support the SQL standard and will it work there or not though.
select mrc, insert_dt from
(
select avg(mrc) over (order by insert_dt, id) mrc, insert_dt,
row_number() over (partition by insert_dt order by id desc) rn
from the_table
) t
where rn = 1
order by insert_dt;
DB-fiddle demo
Update If the_table has no id column then use a CTE to add one.
with t_id as (select *, row_number() over (order by insert_dt) id from the_table)
select mrc, insert_dt from
(
select avg(mrc) over (order by insert_dt, id) mrc, insert_dt,
row_number() over (partition by insert_dt order by id desc) rn
from t_id
) t
where rn = 1
order by insert_dt;
I have a log table like below and want to simplfy it by getting min start date and max end date for consecutive Status values for each Id. I tried many window function combinations but no luck.
This is what I have:
This is what want to see:
This is a typical gaps-and-islands problem. You want to aggregate groups of consecutive records that have the same Id and Status.
No need for recursion, here is one way to solve it using window functions:
select
Id,
Status,
min(StartDate) StartDate,
max(EndDate) EndDate
from (
select
t.*,
row_number() over(partition by id order by StartDate) rn1,
row_number() over(partition by id, status order by StartDate) rn2
from mytable t
) t
group by
Id,
Status,
rn1 - rn2
order by Id, min(StartDate)
The query works by ranking records over two different partitions (by Id, and by Id and Status). The difference between the ranks gives you the group each record belongs to. You can run the subquery independently to see what it returns and understand the logic.
Demo on DB Fiddle:
Id | Status | StartDate | EndDate
-: | :----- | :------------------ | :------------------
1 | B | 07/02/2019 00:00:00 | 18/02/2019 00:00:00
1 | C | 18/02/2019 00:00:00 | 10/03/2019 00:00:00
1 | B | 10/03/2019 00:00:00 | 01/04/2019 00:00:00
2 | A | 05/02/2019 00:00:00 | 22/04/2019 00:00:00
2 | D | 22/04/2019 00:00:00 | 05/05/2019 00:00:00
2 | A | 05/05/2019 00:00:00 | 30/06/2019 00:00:00
Try the following query. First order the data by StartDate and generate a sequence (rid). Then you the recursive cte to get the first row (rid=1) for each group (id,status), and recursively get the next row and compare the start/end date.
;WITH cte_r(id,[Status],StartDate,EndDate,rid)
AS
(
SELECT id,[Status],StartDate,EndDate, ROW_NUMBER() OVER(PARTITION BY Id,[Status] ORDER BY StartDate) AS rid
FROM log_table
),
cte_range(id,[Status],StartDate,EndDate,rid)
AS
(
SELECT id,[Status],StartDate,EndDate,rid
FROM cte_r
WHERE rid=1
UNION ALL
SELECT p.id, p.[Status], CASE WHEN c.StartDate<p.EndDate THEN p.StartDate ELSE c.StartDate END AS StartDate, c.EndDate,c.rid
FROM cte_range p
INNER JOIN cte_r c
ON p.id=c.id
AND p.[Status]=c.[Status]
AND p.rid+1=c.rid
)
SELECT id,[Status],StartDate,MAX(EndDate) AS EndDate FROM cte_range GROUP BY id,StartDate ;
I am playing around with bigquery. Following input is given:
+---------------+---------+---------+--------+----------------------+
| customer | agent | value | city | timestamp |
+---------------+---------+---------+--------+----------------------+
| 1 | 1 | 106 | LA | 2019-02-12 03:05pm |
| 1 | 1 | 251 | LA | 2019-02-12 03:06pm |
| 3 | 2 | 309 | NY | 2019-02-12 06:41pm |
| 1 | 1 | 654 | LA | 2019-02-12 05:12pm |
+---------------+---------+---------+--------+----------------------+
I want to find transactions that where issued one after another (say within 5 minutes) by one and the same agent. So the output for the above table should look like:
+---------------+---------+---------+--------+----------------------+
| customer | agent | value | city | timestamp |
+---------------+---------+---------+--------+----------------------+
| 1 | 1 | 106 | LA | 2019-02-12 03:05pm |
| 1 | 1 | 251 | LA | 2019-02-12 03:06pm |
+---------------+---------+---------+--------+----------------------+
The query should somehow group by agent and find such transactions. However the result is not really grouped as you can see from the output. My first thought was using the LEAD function, but I am not sure. Do you have any ideas?
Ideas for a query:
sort by agent and timestamp DESC
start with the first row, look at the following row (using LEAD?)
check if timestamp difference is less than 5 minutes
if so, this two rows should be in the output
continue with next (2nd) row
When the 2nd and 3rd row match the criteria, too, the 2nd row will get into the output, which would cause duplicate rows. I am not sure how to avoid that, yet.
There must be an easier way but does this achieve what you are after?
CTE2 AS (
SELECT customer, agent, value, city, timestamp,
lead(timestamp,1) OVER (PARTITION BY agent ORDER BY timestamp) timestamp_lead,
lead(customer,1) OVER (PARTITION BY agent ORDER BY timestamp) customer_lead,
lead(value,1) OVER (PARTITION BY agent ORDER BY timestamp) value_lead,
lead(city,1) OVER (PARTITION BY agent ORDER BY timestamp) city_lead,
lag(timestamp,1) OVER (PARTITION BY agent ORDER BY timestamp) timestamp_lag
FROM CTE
)
SELECT agent,
if(timestamp_diff(timestamp_lead,timestamp,MINUTE)<5, concat(cast(customer as string),', ',cast(customer_lead as string)),cast(customer as string)) customer,
if(timestamp_diff(timestamp_lead,timestamp,MINUTE)<5, concat(cast(value as string),', ',cast(value_lead as string)),cast(value as string)) value,
if(timestamp_diff(timestamp_lead,timestamp,MINUTE)<5, concat(cast(city as string),', ',cast(city_lead as string)),cast(city as string)) cities,
if(timestamp_diff(timestamp_lead,timestamp,MINUTE)<5, concat(cast(timestamp as string),', ',cast(timestamp_lead as string)),cast(timestamp as string)) timestamps
FROM CTE2
WHERE (timestamp_diff(timestamp_lead,timestamp,MINUTE)<5 OR NOT timestamp_diff(timestamp,timestamp_lag,MINUTE)<5)
Below is for BigQuery Standard SQL
#standardSQL
SELECT * FROM (
SELECT *,
IF(TIMESTAMP_DIFF(LEAD(ts) OVER(PARTITION BY agent ORDER BY ts), ts, MINUTE) < 5,
LEAD(STRUCT(customer AS next_customer, value AS next_value)) OVER(PARTITION BY agent ORDER BY ts),
NULL).*
FROM `project.dataset.yourtable`
)
WHERE NOT next_customer IS NULL
You can test, play with above using sample data from your question as in below example
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 customer, 1 agent, 106 value,'LA' city, '2019-02-12 03:05pm' ts UNION ALL
SELECT 1, 1, 251,'LA', '2019-02-12 03:06pm' UNION ALL
SELECT 3, 2, 309,'NY', '2019-02-12 06:41pm' UNION ALL
SELECT 1, 1, 654,'LA', '2019-02-12 05:12pm'
), temp AS (
SELECT customer, agent, value, city, PARSE_TIMESTAMP('%Y-%m-%d %I:%M%p', ts) ts
FROM `project.dataset.table`
)
SELECT * FROM (
SELECT *,
IF(TIMESTAMP_DIFF(LEAD(ts) OVER(PARTITION BY agent ORDER BY ts), ts, MINUTE) < 5,
LEAD(STRUCT(customer AS next_customer, value AS next_value)) OVER(PARTITION BY agent ORDER BY ts),
NULL).*
FROM temp
)
WHERE NOT next_customer IS NULL
-- ORDER BY ts
with result
Row customer agent value city ts next_customer next_value
1 1 1 106 LA 2019-02-12 15:05:00 UTC 1 251
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)