Question

我对SQL很新。我的桌子是

   id     mark  datetimes      
    ------|-----|------------
    1001  | 10  | 2011-12-20   
    1002  | 11  | 2012-01-10 
    1005  | 12  | 2012-01-10  
    1003  | 10  | 2012-01-10 
    1004  | 11  | 2018-10-10 
    1006  | 12  | 2018-10-19  
    1007  | 13  | 2018-03-12  
    1008  | 15  | 2018-03-13

我需要在每个月末选择一个标记最高的ID（年份也很重要），ID可以重复

我想要的输出是

    id    mark  
    -----|----
   1001  | 10
   1005  | 12
   1006  | 12
   1008  | 15

到目前为止，我只能在每个月获得最高价值

 Select Max(Mark)'HighestMark' 
 From StudentMark
 Group BY Year(datetimes), Month(datetimes)

当我试图

时

 Select Max(Mark)'HighestMark', ID 
 From StudentMark
 Group BY Year(datetimes), Month(datetimes), ID

我得到了

Id          HighestMark
----------- ------------
1001        10
1002        11
1003        12
1004        10
1005        11
1006        12
1007        13
1008        15

Answer 1

我没有看到在单个查询中执行此操作的方法。但是我们可以很容易地使用一个子查询来为每个学生找到月份的最终分数，而另一个子分数可以找到最终分数最高的学生。

WITH cte AS (
    SELECT *,
        ROW_NUMBER() OVER (PARTITION BY ID, CONVERT(varchar(7), datetimes, 126)
            ORDER BY datetimes DESC) rn
    FROM StudentMark
)

SELECT ID, Mark AS HighestMark
FROM
(
    SELECT *,
        RANK() OVER (PARTITION BY CONVERT(varchar(7), datetimes, 126)
            ORDER BY Mark DESC) rk
    FROM cte
    WHERE rn = 1
) t
WHERE rk = 1
ORDER BY ID;

Demo

Answer 2

您可以尝试以下内容。

使用ROW_NUMBER()

SELECT * FROM
(
  SELECT *, 
   ROW_NUMBER() OVER(PARTITION BY YEAR(DATETIMES)
               ,MONTH(DATETIMES) ORDER BY MARK DESC) AS RN    
 FROM [MY_TABLE]
 )T WHERE RN=1

使用WITH TIES

SELECT TOP 1 WITH TIES ID, mark AS HighestMarks
FROM [MY_TABLE]
ORDER BY ROW_NUMBER() OVER (PARTITION BY YEAR(datetimes)
                     ,MONTH(datetimes) ORDER BY mark DESC)

示例：

WITH MY AS ( SELECT * FROM (VALUES (1001 , 10 , '2011-12-20'), (1002 , 11 , '2012-01-10'), (1005 , 12 , '2012-01-10'), (1003 , 10 , '2012-01-10'), (1004 , 11 , '2018-10-10'), (1006 , 12 , '2018-10-19'), (1007 , 13 , '2018-03-12'), (1008 , 15 , '2018-03-13') ) T( id , mark , datetimes) ) SELECT ID,Mark as HighestMark FROM ( SELECT *, ROW_NUMBER() OVER(PARTITION BY YEAR(DATETIMES),MONTH(DATETIMES) ORDER BY MARK DESC) AS RN FROM MY )T WHERE RN=1

<强>输出：

ID HighestMark 1001 10 1005 12 1008 15 1006 12

Answer 3

如果有超过1名学生具有相同的最高分，则使用RANK。

select id, mark
from
(select *,  
 rank() over( partition by convert(char(7), datetimes, 111) order by mark desc) seqnum
from studentMark ) t
where seqnum = 1

Answer 4

在下面的查询中，您已为分组依据添加了 ID 列，因此，它正在考虑所有ID的所有数据。

Select Max(Mark)'HighestMark', ID From StudentMark Group BY Year(datetimes), Month(datetimes), ID

从此脚本中删除ID列，然后重试。

Answer 5

这应该有效：

    select s.ID, t.Mark, t.[Month year] from Studentmark s
    inner join (
    Select 
Max(Mark)'HighestMark'
,cast(Year(datetimes) as varchar(10)) +  
 cast(Month(datetimes) as varchar(10)) [month year]
     From StudentMark
     Group BY cast(Year(datetimes) as varchar(10)) 
           +  cast(Month(datetimes) as varchar(10))) t on t.HighestMark = s.mark and  
t.[month year] = cast(Year(s.datetimes) as varchar(10)) +  cast(Month(s.datetimes) as varchar(10))

Answer 6

如果出于某种原因你厌恶子查询，你实际上可以这样做：

select distinct
      first_value(id) over (partition by year(datetimes), month(datetime) order by mark desc) as id
      max(mark) over (partition by year(datetimes), month(datetime))
from StudentMark;

或者：

select top (1) with ties id, mark
from StudentMark
order by row_number() over (partition by year(datetimes), month(datetime) order by mark desc);

在这种情况下，您可以使用rank()或dense_rank()代替row_number()让所有学生参与其中。

如何每月选择具有最高分数的ID

6 个答案:

Demo