如何编写sql查询以选择一列中具有最大值的行

Question

我的表看起来像这样。

Id   |  Name  |  Ref  | Date        | From
10   |  Ant   |  100  | 2017-02-02  | David
10   |  Ant   |  300  | 2016-01-01  | David
2    |  Cat   |  90   | 2017-09-09  | David
2    |  Cat   |  500  | 2016-02-03  | David
3    |  Bird  |  150  | 2017-06-28  | David

这是我想要的结果。

Id   |  Name  |  Ref  | Date       | From
3    |  Bird  |  150  | 2017-06-28 | David
2    |  Cat   |  500  | 2016-02-03 | David
10   |  Ant   |  300  | 2016-01-01 | David

我的目标是按订单日期desc排序的最高每个Id的参考。

能否告诉我如何使用pl / sql编写SQL查询。

Answer 1

这种要求（您需要最多或最少一列，按另一列分组，但您需要来自最大或最小行的所有数据）几乎是分析函数的用途。 我使用了row_number - 如果关系是可能的，你需要澄清分配（参见我的问题下的评论），并且根据细节，另一个分析函数可能更合适 - 也许是rank() 。

with
     my_table ( id, name, ref, dt, frm ) as (
       select 10, 'Ant' , 100, date '2017-02-02', 'David' from dual union all
       select 10, 'Ant' , 300, date '2016-01-01', 'David' from dual union all
       select  2, 'Cat' ,  90, date '2017-09-09', 'David' from dual union all
       select  2, 'Cat' , 500, date '2016-02-03', 'David' from dual union all
       select  3, 'Bird', 150, date '2017-06-28', 'David' from dual
     )
-- End of simulated table (for testing purposes only, not part of the solution).
-- SQL query begins BELOW THIS LINE.
select   id, name, ref, dt, frm
from     (
           select id, name, ref, dt, frm,
                  row_number() over (partition by id order by ref desc, dt desc) as rn
           from   my_table
         )
where    rn = 1
order by dt desc
;

ID  NAME  REF  DT          FRM 
--  ----  ---  ----------  -----
 3  Bird  150  2017-06-28  David
 2  Cat   500  2016-02-03  David
10  Ant   300  2016-01-01  David

Answer 2

你可以用它

SELECT
Id
,Name
,Ref
,[Date]
FROM(
SELECT 
* 
, ROW_NUMBER() OVER(PARTITION BY ID ORDER BY Ref DESC) AS Row#
FROM yourtable
) A WHERE Row# = 1
ORDER BY A.[Date] DESC

Answer 3

另一个带有自联接的解决方案（Idea来自这里： 我如何用MAX（列值）选择行，DISTINCT用SQL中的另一列？ ）：

with
    my_table ( id, name, ref, dt, frm ) as (
      select 10, 'Ant' , 100, date '2017-02-02', 'David' from dual union all
      select 10, 'Ant' , 300, date '2016-01-01', 'David' from dual union all
      select 10, 'Ant' , 300, date '2015-01-01', 'David' from dual union all
      select  2, 'Cat' ,  90, date '2017-09-09', 'David' from dual union all
      select  2, 'Cat' , 500, date '2016-02-03', 'David' from dual union all
      select  3, 'Bird', 150, date '2017-06-28', 'David' from dual
    )
-- End of simulated table (for testing purposes only, not part of the solution).
-- SQL query begins BELOW THIS LINE.

select m1.* 
from my_table m1 
left join my_table m2 
on m1.id = m2.id and (
 -- this is basically a comparator: order by ref desc, dt desc
 m1.ref < m2.ref or (
   m1.ref = m2.ref and 
   m1.dt < m2.dt
 )
) where m2.id is null order by m1.dt desc
;

        ID NAME        REF DT        FRM 
---------- ---- ---------- --------- -----
         3 Bird        150 28-JUN-17 David
         2 Cat         500 03-FEB-16 David
        10 Ant         300 01-JAN-16 David

Answer 4

使用“优于”SQL主体：

select a.Id, a.Name, a.Ref, a.Dt, a.frm
from table_name a
left join table_name b on a.id = b.id and b.ref > a.ref -- b.ref > a.ref would make b.ref "better" that a
where b.id is null -- Now check and make sure there is nothing "better"
group by a.id;

Answer 5

SELECT Id, Name, Max(Ref) as Ref, Min(`Date`) as `Date`
From Forge
Group By Id, Name
Order by Min(`Date`) desc;