SQL query to select distinct row with minimum value
SqlDatabaseGreatest N-per-GroupSql Problem Overview
I want an SQL statement to get the row with a minimum value.
Consider this table:
id game point
1 x 5
1 z 4
2 y 6
3 x 2
3 y 5
3 z 8
How do I select the ids that have the minimum value in the point
column, grouped by game? Like the following:
id game point
1 z 4
2 y 5
3 x 2
Sql Solutions
Solution 1 - Sql
Use:
SELECT tbl.*
FROM TableName tbl
INNER JOIN
(
SELECT Id, MIN(Point) MinPoint
FROM TableName
GROUP BY Id
) tbl1
ON tbl1.id = tbl.id
WHERE tbl1.MinPoint = tbl.Point
Solution 2 - Sql
This is another way of doing the same thing, which would allow you to do interesting things like select the top 5 winning games, etc.
SELECT *
FROM
(
SELECT ROW_NUMBER() OVER (PARTITION BY ID ORDER BY Point) as RowNum, *
FROM Table
) X
WHERE RowNum = 1
You can now correctly get the actual row that was identified as the one with the lowest score and you can modify the ordering function to use multiple criteria, such as "Show me the earliest game which had the smallest score", etc.
Solution 3 - Sql
This will work
select * from table
where (id,point) IN (select id,min(point) from table group by id);
Solution 4 - Sql
As this is tagged with sql
only, the following is using ANSI SQL and a window function:
select id, game, point
from (
select id, game, point,
row_number() over (partition by game order by point) as rn
from games
) t
where rn = 1;
Solution 5 - Sql
Ken Clark's answer didn't work in my case. It might not work in yours either. If not, try this:
SELECT *
from table T
INNER JOIN
(
select id, MIN(point) MinPoint
from table T
group by AccountId
) NewT on T.id = NewT.id and T.point = NewT.MinPoint
ORDER BY game desc
Solution 6 - Sql
SELECT DISTINCT
FIRST_VALUE(ID) OVER (Partition by Game ORDER BY Point) AS ID,
Game,
FIRST_VALUE(Point) OVER (Partition by Game ORDER BY Point) AS Point
FROM #T
Solution 7 - Sql
SELECT * from room
INNER JOIN
(
select DISTINCT hotelNo, MIN(price) MinPrice
from room
Group by hotelNo
) NewT
on room.hotelNo = NewT.hotelNo and room.price = NewT.MinPrice;
Solution 8 - Sql
This alternative approach uses SQL Server's OUTER APPLY
clause. This way, it
- creates the distinct list of games, and
- fetches and outputs the record with the lowest point number for that game.
The OUTER APPLY
clause can be imagined as a LEFT JOIN
, but with the advantage that you can use values of the main query as parameters in the subquery (here: game).
SELECT colMinPointID
FROM (
SELECT game
FROM table
GROUP BY game
) As rstOuter
OUTER APPLY (
SELECT TOP 1 id As colMinPointID
FROM table As rstInner
WHERE rstInner.game = rstOuter.game
ORDER BY points
) AS rstMinPoints
Solution 9 - Sql
This is portable - at least between ORACLE and PostgreSQL:
select t.* from table t
where not exists(select 1 from table ti where ti.attr > t.attr);
Solution 10 - Sql
Most of the answers use an inner query. I am wondering why the following isn't suggested.
select
*
from
table
order by
point
fetch next 1 row only // ... or the appropriate syntax for the particular DB
This query is very simple to write with JPAQueryFactory
(a Java Query DSL class).
return new JPAQueryFactory(manager).
selectFrom(QTable.table).
setLockMode(LockModeType.OPTIMISTIC).
orderBy(QTable.table.point.asc()).
fetchFirst();
Solution 11 - Sql
Try:
select id, game, min(point) from t
group by id