sql - 返回每个 ID SQL 的最小行数

Question

我有一个很长的列表，看起来像下面这样（列名和数据匿名）

ID | ID2 | Value
1  |  5  |  900
1  |  7  |  400
2  |  2  |  100
2  |  8  |  800
2  |  2  |  200
3  |  4  |  100
3  |  5  |  300
4  |  8  |  750
4  |  5  |  900

我所追求的是能够每列“ID”制作 1 行，它具有最低 ID2 的值的总和。

例如，上表将变为：

ID | ID2 | Value
1  |  5  |  900
2  |  2  |  300
3  |  4  |  100
4  |  5  |  900

我已经尝试了几个脚本变体，但没有成功。这是我当前清理的代码：

    SELECT res.id
       , min(id2) as rdk
       ,sum(value)

  FROM x as res

    left join y as clai
    on res.id = clai.id

    left join z as cal
    on clai.col99 = cal.col99

    group by
       res.id

然而，总和是将“ID”的所有价值相加，而不仅仅是最低的“ID2”？

任何帮助表示赞赏！

Answer 1

尝试这个：-

select a.ID,a.ID2, sum(Value) as Value
from
your_table_name a
inner join
(
SElect ID, min(ID2) as ID2
from your_table_name
group by ID
) b
on a.ID=b.ID and a.ID2=b.ID2
group by a.ID,a.ID2;

希望这可以帮助：-）

Answer 2

如果您的表称为 t1：

SELECT t1.id, t1.id2, SUM(t1.value) 
FROM t1 INNER JOIN
(
 SELECT id, min(id2) minId 
 FROM t1
 GROUP BY id
) A
ON A.id = t1.id AND A.minId = t1.id2
GROUP BY t1.id, t1.id2

Answer 3

我将把我的解决方案留在这里:)

with ExampleData as (
  select 1 ID1,  5 ID2,   900 Value union all
  select 1 ID1,  7  ID2,  400 Value union all
  select 2 ID1,  2  ID2,  100 Value union all
  select 2 ID1,  8  ID2,  800 Value union all
  select 2 ID1,  2  ID2,  200 Value union all
  select 3 ID1,  4  ID2,  100 Value union all
  select 3 ID1,  5  ID2,  300 Value union all
  select 4 ID1,  8  ID2,  750 Value union all
  select 4 ID1,  5  ID2,  900 Value)

select ID1,
       ID2, 
       Value 
from (
   select ID1,
          ID2,
          SUM(VALUE) VALUE,
          ROW_NUMBER() OVER (partition by ID1 order by ID2) R 
   from ExampleData
   group by ID1,ID2
) T where T.R=1

Answer 4

试试这个：

select a.*,b.value from
(select id,min(id2) as id2 
from have
group by id) a
left join
(select id,id2,sum(value) as value
from have
group by id,id2) b
on a.id=b.id and a.id2=b.id2;

我的输出：

id  |id2 |value
1   |5   |900
2   |2   |300
3   |4   |100
4   |5   |900

如有任何疑问，请告诉我。