按两列分组，取总和，然后取最大值

Question

共有三列： Id (char)、 Name (char) 和Score (int)。

首先，我们按Id和Name分组，并为每个组添加Score 。 让我们将添加的分数total_score 。

然后，我们按Name分组，只取total_score及其对应的Id和Name的最大值。 我拥有其他一切，但我很难弄清楚如何获得Id 。 我得到的错误是

select 列表中的“Id”列无效，因为它不包含在聚合 function 或 GROUP BY 子句中。

WITH Tmp AS
  (SELECT Id,
          Name,
          SUM(Score) AS total_score
   FROM Mytable
   GROUP BY Id,
            Name)
SELECT Name,  -- Id,
       MAX(total_score) AS max_score
FROM Tmp
GROUP BY Name
ORDER BY max_score DESC

Answer 1

WITH Tmp AS
  (SELECT Id,
          Name,
          SUM(Score) AS total_score
   FROM Mytable
   GROUP BY Id,
            Name)
SELECT Name,   Id,
       MAX(total_score) AS max_score
FROM Tmp
GROUP BY Name,id
ORDER BY max_score DESC

尝试这个。 希望这会有所帮助。

Answer 2

我不确定以下查询的性能，但我们可以使用window functions从数据partition中获取maximum 。

SELECT
        Id,
        Name,
        SUM(Score) AS total_score,
        MAX(SUM(Score)) OVER(Partition by Name) AS max_score
FROM Mytable
GROUP BY Id, Name;

经测试——

declare @Mytable table (id int, name varchar(10), score int);
insert into @Mytable values
(1,'abc', 100),
(2,'abc', 200),
(3,'def', 300),
(3,'def', 400),
(4,'pqr', 500);

Output -

Id  Name   total_score  max_score
1   abc    100          200
2   abc    200          200
3   def    700          700
4   pqr    500          500

Answer 3

WITH Tmp AS
 (
 SELECT Id,
          Name,
          SUM(Score) AS total_score
   FROM Mytable
   GROUP BY Id,
            NAME
 )
SELECT Name,   Id,
       MAX(total_score) AS max_score
FROM Tmp
GROUP BY Name,id
ORDER BY max_score DESC

注意：- 如果我们使用聚合 function 那么我们必须使用其他列作为 Group By....

在您的情况下，您使用 SUM(Score) 作为聚合 function 然后我们使用其他列作为 Group by...

Answer 4

只需按Name将row_number()分区添加到您的查询中并获取第一行（按total_score降序排列）

select  *
from
(
    -- your existing `total_score` query
    SELECT  Id, Name,
            SUM(Score) AS total_score,
            r = row_number() over (partition by Name order by SUM(Score) desc)
    FROM  Mytable
    GROUP BY Id, Name
) d
where   r = 1

Answer 5

您可以 select DENSE_RANK()使用total_score列，然后使用 select 记录Rank = 1 。 当有多个Name具有相同的total_score时，这也适用于那些。

WITH Tmp AS
  (SELECT Id,
          Name,
          SUM(Score) AS total_score
   FROM Mytable
   GROUP BY Id, Name)
SELECT Id, 
       Name,
       total_score AS max_score
FROM (SELECT Id, 
           Name,
           total_score,
           DENSE_RANK() OVER (PARTITION BY Name ORDER BY total_score DESC) AS Rank
    FROM Tmp) AS Tmp2
WHERE Rank = 1

Answer 6

你也可以试试这个：

select id,name,max(total_score) over (partition by name)  max_score from (
select id,name,sum(score) as total_score from YOURTABLE
group by id,name
) t