选择其他列为组的最大值的列值

Question

我试图在表中选择两列(ID和状态).该表应显示每个ID的最大值的状态.我尝试过其他几个例子但似乎没什么用.

原始数据结构:

ID  state    value (FLOAT)
1   TX   921,294,481 
1   SC   21,417,296 
1   FL   1,378,132,290 
1   AL   132,556,895 
1   NC   288,176 
1   GA   1,270,986,631 
2   FL   551,374,452 
2   LA   236,645,530 
2   MS   2,524,536,050 
2   AL   4,128,682,333 
2   FL   1,503,991,028

因此,结果数据结构应如下所示:

ID  STATE (Max Value)
1   FL
2   AL

佛罗里达州和阿拉巴马州的ID群体价值最高.

任何帮助将不胜感激.我确实在这里找到了答案,但无法让答案对我有用.

Answer 1

对于SQL Server(以及具有窗口函数的其他产品):

SELECT *
FROM
(
   SELECT
     *,
     ROW_NUMBER() OVER (PARTITION BY ID ORDER BY value desc) as rn
   FROM
     UnnamedTable
) t
WHERE
   t.rn = 1

Answer 2

基于value数字假设的解决方案:

SELECT
  [ID],
  [State],
  [Value]
FROM
(
  SELECT 
    [ID],
    [State],
    [Value],
    Rank() OVER (PARTITION BY [ID] ORDER BY [Value] DESC) AS [Rank]
  FROM [t1]
) AS [sub]
WHERE [sub].[Rank] = 1
ORDER BY
  [ID] ASC,
  [State] ASC

如果State具有相同的多个s具有ID相同的Value,则它们都将获得相同的结果Rank.这与使用不同Row_Number,后者返回唯一的行号,但是顺序是任意选择的.(另请参阅:SQL RANK()与ROW_NUMBER())

Answer 3

您可以使用子查询来获得此结果:

select t1.id, t1.[state] MaxValue
from yourtable t1
inner join
(
  select id, max(value) MaxVal
  from yourtable
  group by id
) t2
  on t1.id = t2.id
  and t1.value = t2.maxval
order by t1.id

请参阅SQL Fiddle with Demo