加入并选择具有最大值的列

Question

我正在处理 Oracle 中的两个表，TABLE_A 包含唯一 ID 和余额（数字）。 TABLE_B 显示相同 ID 的特定交易并包含以下字段：ID（非唯一）、BAL、Sequence_number。 我想检查 TABLE_A.Balance 是否始终等于具有 Max(Sequence_number) 的 TABLE_B.Balance。 所以我希望每个 ID 只有一行。

我尝试了以下方法，但它不会为每个 ID 返回唯一的行，而是返回多个。这是为什么呢？

Select a.ID, a.Balance,b.Balance, b.sequence_number From TABLE_A a Inner join (select ID, Balance, max(sequence_number) as sequence_number from TABLE_B group by ID, Balance) b On a.ID = B.ID Group by a.ID, a.Balance, b.Balance, b.sequence_number

TABLE_A

ID_______Balance
1_______10
2_______15
3_______50

TABLE_B

ID____Balance____Sequence_number
1_______19_______1
1_______75_______2
1_______10_______3
2_______39_______1
2_______15_______2
3_______120_______1
3_______89_______2
3_______57_______3
3_______50_______4

Answer 1

你可以使用窗口函数：

select a.*, b.balance
from a left join
     (select b.*,
             row_number() over (partition by id order by sequence_number desc) as seqnum
      from b
     ) b
     on b.id = a.id and b.seqnum = 1;

我不太确定您要比较什么，但这会返回a 中的每一行以及b 中具有最高序列号的行。

Answer 2

您可以使用 row_number()over() 窗口函数来获取 ID 明智的最高序列号的行。

架构：

create table TABLE_A(ID int, Balance int);
insert into TABLE_A values(1,10);
insert into TABLE_A values(2,15);
insert into TABLE_A values(3,50);

create table TABLE_B (ID int , Balance int,Sequence_number int);
insert into TABLE_B values(1,19,1);
insert into TABLE_B values(1,75,2);
insert into TABLE_B values(1,10,3);
insert into TABLE_B values(2,39,1);
insert into TABLE_B values(2,15,2);
insert into TABLE_B values(3,120,1);
insert into TABLE_B values(3,89,2);
insert into TABLE_B values(3,57,3);
insert into TABLE_B values(3,50,4);
GO

查询：

    with cte as 
               (
               select id, balance, sequence_number, row_number()over (partition by id order by sequence_number desc)rnk from table_b
               )
    Select a.ID, a.Balance,b.Balance,  b.sequence_number From TABLE_A a  
    Inner join cte b on a.id=b.id and rnk=1;
GO

输出：

ID	Balance	Balance	sequence_number
1	10	10	3
2	15	15	2
3	50	50	4

db小提琴here