按 ID 和 select MAX 分组

Question

晚上好，我正在 Oracle 中处理这样的 table：

ID	BALANCE	SEQ
1	102	13
1	119	15
2	50	4
3	20	11
3	15	10
3	45	9
4	90	5
5	67	20
5	12	19
6	20	1

我想select，对于每个 ID，具有 MAX(SEQ) 的 BALANCE。

所以最终结果将是：

ID	BALANCE	SEQ
1	119	15
2	50	4
3	20	11
4	90	5
5	67	20
6	20	1

我该怎么做？我尝试了几个 Group by queries 但没有成功。感谢您的帮助

Answer 1

一种方法是使用 keep 聚合：

select id,
       max(balance) keep (dense_rank first order by seq desc) as balance,
       max(seq)
from t
group by id;

Answer 2

你可以使用普通的rank()

SELECT ID, BALANCE, SEQ FROM ( 
select 
ID, BALANCE, SEQ, RANK() OVER (PARTITION BY ID ORDER BY SEQ DESC)  ranks
from t 
) WHERE ranks = 1

示例演示

    SELECT ID, BALANCE, SEQ FROM ( 
SELECT ID, BALANCE, SEQ, RANK() OVER (PARTITION BY ID ORDER BY SEQ DESC)  ranks 
FROM ( 
SELECT 1 ID,    102 BALANCE,    13 SEQ FROM dual UNION all
SELECT 1,       119,    15  FROM dual UNION all
SELECT 2,       50,     4   FROM dual UNION all
SELECT 3,       20,     11  FROM dual UNION all
SELECT 3,       15,     10  FROM dual UNION all
SELECT 3,       45,     9   FROM dual UNION all
SELECT 4,       90,     5   FROM dual UNION all
SELECT 5,       67,     20  FROM dual UNION all
SELECT 5,       12,     19  FROM dual UNION all
SELECT 6,       20,     1   FROM dual 
)
) WHERE ranks = 1

您可以将其添加到您的大查询中，如下所示

SELECT ID, BALANCE, SEQ FROM ( 
select 
ID, BALANCE, SEQ, RANK() OVER (PARTITION BY ID ORDER BY SEQ DESC)  ranks
from (**YOUR BIG QUERY HERE**) 
) WHERE ranks = 1

按 ID 和 select MAX 分组

GROUP BY ID and select MAX

sql

oracle

group-by

max

seq