Question

如果我有下表：

Table name: FOO
ID | NUMBER | EVENT | NAME | CAR
1  |   12   | OFFER | Adam | VW
2  |   13   | ORDER | Adam | VW
3  |   11   | OFFER | Adam | BMW
4  |   12   | OFFER | Adam | BMW
5  |   3    | OFFER | Adam | BMW
6  |   1    | ORDER | Mark | Mercedes
7  |   1    | ORDER | Mark | BMW

查询应该返回

Result name: FOO
ID | NUMBER | EVENT | NAME | CAR
2  |   13   | ORDER | Adam | VW
4  |   12   | OFFER | Adam | BMW
6  |   1    | ORDER | Mark | Mercedes
7  |   1    | ORDER | Mark | BMW

以下规则适用：

- Get the records which have DISTINCT NAME AND CAR
- If more than one record with the same NAME AND CAR exists, get the one with MAX NUMBER
- ID, NUMBER, EVENT should not count towards the DISTINCT row rule

数据库平台是DB2，但只要它是纯SQL，任何SQL都是好的。

我无法在整行上将DISTINCT规则仅应用于（NAME，CAR），如果存在多个相同的行（NAME，CAR），则获取NUMBER最高的行。

Answer 1

纯粹的，纯粹的ansi-92 SQL：

select t1.*
from Foo t1
inner join
(
select Name, Car, max(number) as maxNo
from Foo
group by Name, Car
) x2
on t1.Name = x2.Name
and t1.Car = x2.car
and t1.Number = x2.maxNo

Answer 2

这应该可以解决问题

with temp as (
select id, number, event, name, car,
       rownumber() over (partition by name, car order by number desc) as rownum
  from foo
 )
 select id, number, event, name, car from temp 
 where rownum = 1

Answer 3

我认为一个小组可以提供帮助。

select id, max(number), event, name, car from foo group by name, car

这适用于MySQL，但不适用于DB。显然，MySQL更直接。

Answer 4

尝试此SELECT语句

SELECT FOO.ID, F.Number, Foo.Event, f.Name, f.Car
FROM FOO INNER JOIN
(
    SELECT MAX(Number) AS Number, Name, Car 
    FROM FOO
    GROUP BY Name, Car
) F ON FOO.Number = F.Number AND FOO.Name = F.Name AND Foo.Car = F.Car
ORDER BY FOO.ID

查询仅在列组

4 个答案: