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
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`
其他回答
您可以连接一个取出MAX(Group)和Age的子查询。这个方法在大多数RDBMS中是可移植的。
SELECT t1.*
FROM yourTable t1
INNER JOIN
(
SELECT `Group`, MAX(Age) AS max_age
FROM yourTable
GROUP BY `Group`
) t2
ON t1.`Group` = t2.`Group` AND t1.Age = t2.max_age;
我的解决方案只适用于你只需要检索一个列，但我的需求是在性能方面发现的最佳解决方案(它只使用一个查询!):
SELECT SUBSTRING_INDEX(GROUP_CONCAT(column_x ORDER BY column_y),',',1) AS xyz,
column_z
FROM table_name
GROUP BY column_z;
它使用GROUP_CONCAT以创建一个有序concat列表，然后我只将子字符串字符串到第一个。
在PostgreSQL中，你可以使用DISTINCT ON子句:
SELECT DISTINCT ON ("group") * FROM "mytable" ORDER BY "group", "age" DESC;
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;
