如何获得包含每个分组集的最大值的行?

我见过这个问题的一些过于复杂的变体,没有一个有好的答案。我试着把最简单的例子放在一起:

给一个这样的表,有个人、组和年龄列,你如何得到每组中年龄最大的人?(一组中平局的结果应按首字母顺序排列)

Person | Group | Age
---
Bob  | 1     | 32  
Jill | 1     | 34  
Shawn| 1     | 42  
Jake | 2     | 29  
Paul | 2     | 36  
Laura| 2     | 39  

期望结果集:

Shawn | 1     | 42    
Laura | 2     | 39  

当前回答

使用CTEs -常用表表达式:

WITH MyCTE(MaxPKID, SomeColumn1)
AS(
SELECT MAX(a.MyTablePKID) AS MaxPKID, a.SomeColumn1
FROM MyTable1 a
GROUP BY a.SomeColumn1
  )
SELECT b.MyTablePKID, b.SomeColumn1, b.SomeColumn2 MAX(b.NumEstado)
FROM MyTable1 b
INNER JOIN MyCTE c ON c.MaxPKID = b.MyTablePKID
GROUP BY b.MyTablePKID, b.SomeColumn1, b.SomeColumn2

--Note: MyTablePKID is the PrimaryKey of MyTable

其他回答

使用CTEs -常用表表达式:

WITH MyCTE(MaxPKID, SomeColumn1)
AS(
SELECT MAX(a.MyTablePKID) AS MaxPKID, a.SomeColumn1
FROM MyTable1 a
GROUP BY a.SomeColumn1
  )
SELECT b.MyTablePKID, b.SomeColumn1, b.SomeColumn2 MAX(b.NumEstado)
FROM MyTable1 b
INNER JOIN MyCTE c ON c.MaxPKID = b.MyTablePKID
GROUP BY b.MyTablePKID, b.SomeColumn1, b.SomeColumn2

--Note: MyTablePKID is the PrimaryKey of MyTable

让桌子的名字是人

select O.*              -- > O for oldest table
from people O , people T
where O.grp = T.grp and 
O.Age = 
(select max(T.age) from people T where O.grp = T.grp
  group by T.grp)
group by O.grp; 
with CTE as 
(select Person, 
[Group], Age, RN= Row_Number() 
over(partition by [Group] 
order by Age desc) 
from yourtable)`


`select Person, Age from CTE where RN = 1`

不确定MySQL是否有row_number函数。如果是这样,您可以使用它来获得所需的结果。在SQL Server上,你可以做类似的事情:

CREATE TABLE p
(
 person NVARCHAR(10),
 gp INT,
 age INT
);
GO
INSERT  INTO p
VALUES  ('Bob', 1, 32);
INSERT  INTO p
VALUES  ('Jill', 1, 34);
INSERT  INTO p
VALUES  ('Shawn', 1, 42);
INSERT  INTO p
VALUES  ('Jake', 2, 29);
INSERT  INTO p
VALUES  ('Paul', 2, 36);
INSERT  INTO p
VALUES  ('Laura', 2, 39);
GO

SELECT  t.person, t.gp, t.age
FROM    (
         SELECT *,
                ROW_NUMBER() OVER (PARTITION BY gp ORDER BY age DESC) row
         FROM   p
        ) t
WHERE   t.row = 1;

采用排名法。

SELECT @rn :=  CASE WHEN @prev_grp <> groupa THEN 1 ELSE @rn+1 END AS rn,  
   @prev_grp :=groupa,
   person,age,groupa  
FROM   users,(SELECT @rn := 0) r        
HAVING rn=1
ORDER  BY groupa,age DESC,person

这个sql可以解释如下:

Select * from users, (Select @rn:= 0) r 按组别、年龄、个人排序 @prev_grp为空 @rn:= CASE WHEN @prev_grp <> groupa THEN 1 ELSE @rn+1 END 这是一个三个运算符的表达式 就像这样,rn= 1 if prev_grp != groupa else rn=rn+1 让rn=1过滤掉你需要的行